PHP如何将键和值添加到关联数组答案

【问题标题】：PHP how to add keys and values to an associative arrayPHP如何将键和值添加到关联数组
【发布时间】：2015-09-18 23:42:54
【问题描述】：

我有一个关于关联数组的问题，由于缺乏编程知识，我无法自己解决。

我有一个从数字“1”到“12”的 foreach 循环。对于每个数字，我都会获得一个从另一台服务器提取数据的链接。

此数据具有 json 数组的形式，因此我使用 CURL 为从 json 数据（例如雇主名称；雇主 ID；雇主工资；...）中提取的每个键值类型创建单个数组。

我需要将所有这些数据插入到数据库中，因此我创建了一个包含所有这些数据的数组，然后将所有这些数据插入到表中。但是......如果我必须从第一个 foreach 创建最终的单个数组，我没有问题，但我无法将已经存在的最终数组从第二个 foreach 继续到最后一个...... 我需要这样工作的原因可能是从 Json 数组中提取的数据包含相同的 ex。雇主两次，所以我想创建没有重复的最终数组。

代码如下：

createUrl();

function createUrl() {

$serverName = 'test';
$iscountryID = 'test';
$countryCurrency = 'test';
$countryName = 'test';

// TAX url (here I get tax data)*******************************
$urlTaxcountry =  strtolower("https://www.test.org/".$serverName."/tax/".$iscountryID.".json");

// CURRENCY EXCHANGE url **********************
$urlChangeCurrencySell = strtolower("https://www.test.org/".$serverName."/exchange/gold-".$countryCurrency.".json"); // to change currency value in gold    

$Skill = array(1,2,3,4,5,6,7,8,9,10,11,12);

foreach ($Skill as $workSkill) {
   // JOBS url *******************************
   $urlJobcountry = strtolower("https://www.test.org/".$serverName."/job/".$iscountryID."-".$workSkill.".json"); 

  getJsonData($serverName,$iscountryID,$workSkill,$urlJobcountry,$urlChangeCurrencySell,$urlTaxcountry,$countryCurrency,$countryName);
}};

 function getJsonData($serverName,$iscountryID,$workSkill,$urlJobcountry,$urlChangeCurrencySell,$urlTaxcountry,$countryCurrency,$countryName)
    {


    $arrEmpName = array();      
    $arrEmpID = array();
    $arrProduct = array();
    $arrQuality = array();
    $arrSkill = array();
    $arrSalary = array();
    $arrKeys = array();
    $i = 0;

    if ($iscountryID != NULL) {
    // FOR JOBS         
    $chJob = curl_init($urlJobcountry);
    curl_setopt($chJob, CURLOPT_RETURNTRANSFER, true);
    $resultJob = curl_exec($chJob);
    $outLoopJob = json_decode($resultJob, true);

    $marketCountryID = $outLoopJob['country']['id'];
    $marketCountryName = $outLoopJob['country']['name'];

    $f = 0;

    $dataJob = $outLoopJob['offers'];
    foreach($dataJob as $job) 
            { 
                $empCompanyID = ($job['company']['id']);
                $empCompanyName = ($job['company']['name']);
                $empProduct = ($job['product']);
                $empQuality = ($job['quality']);
                $empSkill = ($job['skill']);
                $empSalary = ($job['salary']);  

                // CHECK
                echo "<br>empCompanyID: ".$empCompanyID;
                echo "<br>empCompanyName: ".$empCompanyName;
                echo "<br>empProduct: ".$empProduct;
                echo "<br>empQuality: ".$empProduct;
                echo "<br>empSkill: ".$empSkill;
                echo "<br>empSalary: ".$empSalary;

                $arrEmpName[] = $empCompanyName;        
                $arrEmpID[] = $empCompanyID;
                $arrProduct[] = $empProduct;
                $arrQuality[] = $empQuality;
                $arrSkill[] = $empSkill;
                $arrSalary[] = $empSalary;
                $arrKeys[] = $i++;
                $r++;
            }}

        //$result = array_merge($arrKeys, $arrEmpName, $arrEmpID, $arrProduct, $arrQuality, $arrSkill, $arrSalary); 

        $allJobs = array();

        foreach ($arrKeys as $id => $key) {
            if (empty($allJobs)) {
                $allJobs[$key] = array(
                'EmployerName' => $arrEmpName[$id],
                'EmployerID' => $arrEmpID[$id],
                'Product' => $arrProduct[$id],
                'Quality' => $arrQuality[$id],
                'Skill' => $arrSkill[$id],
                'Salary' => $arrSalary[$id],);
            } elseif ((!in_array($empCompanyID, $allJobs)) && (!in_array($empCompanyName, $allJobs)) && (!in_array($empProduct, $allJobs)) && (!in_array($empQuality, $allJobs)) && (!in_array($empSkill, $allJobs)) && (!in_array($empSalary, $allJobs))) {

比我尝试过的：

                $allJobs[$key] = array(
                'EmployerName' => $arrEmpName[$id],
                'EmployerID' => $arrEmpID[$id],
                'Product' => $arrProduct[$id],
                'Quality' => $arrQuality[$id],
                'Skill' => $arrSkill[$id],
                'Salary' => $arrSalary[$id]);

比

                array_push($allJobs, ("'EmployerName' => ".$arrEmpName[$id]), ("'EmployerID' => ".$arrEmpID[$id]), "'Product' => ".$arrProduct[$id], "'Quality' => ".$arrQuality[$id], "'Skill' => ".$arrSkill[$id], "'Salary' => ".$arrSalary[$id]);
                $f++;
            }};

两者都没有工作... 比代码这样结束：

        // se no entry, write no record
        if  ($empCompanyID == NULL)
            {
                $marketCountryName = $countryName;  // NBBBBBBBBBB
                $empCompanyName = 'No Record';
                $empProduct = 'No Record';
                $empQuality = 'No Record';
                $empSkill = 'No Record';
                $empSalary = 'No Record';
            }; 
        }


        echo "<br>this is allJobs:<br>";
        print_r($allJobs);
        echo "<br>";
        print_r($result);   
   };

感谢您的时间和帮助。

更新

我们开始：这是我从 Json 文件中获取的数据

{"country":
{"id":24
,"name":"Argentina"
}
,"offers":[
    {"company":
        {"id":12767
            ,"name":"Armas Q5 Argentina "
        }
        ,"product":"Weapon"
        ,"quality":5
        ,"skill":8
        ,"salary":25
    }
    ,{"company":
        {"id":12767
            ,"name":"Armas Q5 Argentina "
        }
        ,"product":"Weapon"
        ,"quality":5
        ,"skill":1
        ,"salary":25
    }
    ,{"company":
        {"id":16587
             ,"name":"International bank of Railman"
        }
        ,"product":"Iron"
        ,"quality":1
        ,"skill":1
        ,"salary":1
    }
]
,"load":1.4469039440155
}

此特定数据来自与阿根廷文件对应的文件，工作技能为 8。您必须检查所有技能级别才能获得从 1 到 12 的完整就业市场数据。每个页面仅返回 15 个结果，并且可能会发生“公司 xx”也出现在 Skill Work ex 的页面中。 “3”和“2”。

我会尝试从所有工作技能页面（从 1 到 12）中提取数据并创建一个没有重复的数组，例如：

  [0] => ['EmployerName'] => test ['EmployerID'] => 18768 ['Product'] => test ['Quality'] => 5 ['Skill'] => 12 ['Salary'] => 35.5 ) 
  [1] => ['EmployerName'] => test ['EmployerID'] => 10844 ['Product'] => test ['Quality'] => 5 ['Skill'] => 12 ['Salary'] => 35.01)
  [2] => ['EmployerName'] => ...

更新 2

但如果我只使用：

  } elseif ((!in_array($empCompanyID, $allJobs)) &&   (!in_array($empCompanyName, $allJobs)) && (!in_array($empProduct, $allJobs)) && (!in_array($empQuality, $allJobs)) && (!in_array($empSkill, $allJobs)) && (!in_array($empSalary, $allJobs))) {
                $allJobs[$key] = array(
                'EmployerName' => $arrEmpName[$id],
                'EmployerID' => $arrEmpID[$id],
                'Product' => $arrProduct[$id],
                'Quality' => $arrQuality[$id],
                'Skill' => $arrSkill[$id],
                'Salary' => $arrSalary[$id]);
                $f++;
            }};

我得到了这个奇怪的结果...：

对不起，但我真的不明白为什么......

更新 3

以下是我在示例中使用的 2 个“allJobs 数组”示例：第一个是从 SKILL PAGE 2 获得的（所以是循环的第二个值），第二个是从 SKILL PAGE 3 获得的

这是所有工作（工作技能 2）：

 Array ( [0] => Array ( [EmployerName] => Air Veneto Q1 -1 [EmployerID] => 3481 [Product] => Iron [Quality] => 2 [Skill] => 2 [Salary] => 12.01 ) [1] => Array ( [EmployerName] => Israel Weap Q3 [EmployerID] => 2533 [Product] => Weapon [Quality] => 3 [Skill] => 2 [Salary] => 11 ) [2] => Array ( [EmployerName] => Israel Weap Q3 [EmployerID] => 2533 [Product] => Weapon [Quality] => 3 [Skill] => 1 [Salary] => 10.61 ) [3] => Array ( [EmployerName] => Traforis [EmployerID] => 15848 [Product] => Oil [Quality] => 4 [Skill] => 2 [Salary] => 9.05 ) [4] => Array ( [EmployerName] => 1weapQ5 [EmployerID] => 10431 [Product] => Weapon [Quality] => 5 [Skill] => 1 [Salary] => 7.2 ) [5] => Array ( [EmployerName] => WAR 1111 [EmployerID] => 17753 [Product] => Iron [Quality] => 5 [Skill] => 2 [Salary] => 7.1 ) [6] => Array ( [EmployerName] => Silent murder [EmployerID] => 16005 [Product] => Weapon [Quality] => 1 [Skill] => 1 [Salary] => 6.5 ) [7] => Array ( [EmployerName] => NK Iron [EmployerID] => 16745 [Product] => Iron [Quality] => 3 [Skill] => 2 [Salary] => 6.5 ) [8] => Array ( [EmployerName] => NK Iron [EmployerID] => 16745 [Product] => Iron [Quality] => 3 [Skill] => 1 [Salary] => 5 ) [9] => Array ( [EmployerName] => Air Veneto Q1 -1 [EmployerID] => 3481 [Product] => Iron [Quality] => 2 [Skill] => 1 [Salary] => 5 ) [10] => Array ( [EmployerName] => Helvetia Eisen [EmployerID] => 10673 [Product] => Iron [Quality] => 4 [Skill] => 1 [Salary] => 3 ) [11] => Array ( [EmployerName] => Olimpiaki A.E. [EmployerID] => 7156 [Product] => Ticket [Quality] => 1 [Skill] => 1 [Salary] => 1.01 ) [12] => Array ( [EmployerName] => Africa Oil Corp. [EmployerID] => 18180 [Product] => Oil [Quality] => 2 [Skill] => 1 [Salary] => 1 ) [13] => Array ( [EmployerName] => Hardika Olaj #1 [EmployerID] => 7861 [Product] => Oil [Quality] => 3 [Skill] => 1 [Salary] => 1 ) [14] => Array ( [EmployerName] => GD & Co. [EmployerID] => 18770 [Product] => Weapon [Quality] => 1 [Skill] => 1 [Salary] => 1 ) )

这是所有工作（工作技能 3）：

如你所见，

  Array ( [EmployerName] => Air Veneto Q1 -1 [EmployerID] => 3481 [Product] => Iron [Quality] => 2 [Skill] => 2 [Salary] => 12.01

重复两次，两个条目完全相同。

更新 4（解决方案）

好吧，我无法解决关联数组的问题，所以（这可能对其他人有用，我稍后在代码中进行了检查，就在将数据插入数据库之前。

所以当我发现所有 VARS 已经作为存储在 DB 中的值存在时，代码会跳转到下一部分，避免再次添加相同的值。

代码如下：

 function addJobs($iscountryID, $countryName, $countryCurrency,  $empCompanyID, $empCompanyName, $empProduct, $empQuality, $empSkill, $empSalary, $Changerate, $goldJob, $taxIncome, $netgoldJob, $serverName, $linktopageJob, $linktoEmployer, $dayTimeJob, $timeJob, $Currencysell, $marketCountryID, $marketCountryName)

{

 // db connection
 require "db-connection/db-connection2.php";

 $sqlfind = "SELECT 
 ID 
 FROM cJM".$serverName." 
 WHERE (
 iscountryID = ? 
 AND countryName = ? AND countryCurrency = ? 
 AND empCompanyID = ? AND empCompanyName = ? 
 AND empProduct = ? AND empQuality = ? 
 AND empSkill = ? AND empSalary = ? 
 AND goldExchange = ? AND salaryinGold = ? 
 AND jobTax = ? AND netSalary = ? AND Server = ?
 AND time_of_record = ? AND DateSave = CURDATE()
 )";

 echo "<b><br><br>*********************POST***********************</b>";
 echo "<br>empCompanyID: ".$empCompanyID;
 echo "<br>empCompanyName: ".$empCompanyName;
 echo "<br>empProduct: ".$empProduct;
 echo "<br>empQuality: ".$empProduct;
 echo "<br>empSkill: ".$empSkill;
 echo "<br>empSalary: ".$empSalary;                 
 echo "<br>is country id: ".$iscountryID;
 echo "<br>country name: ".$countryName;
 echo "<br>country currency: ".$countryCurrency;
 echo "<br>gold exchange rate: ".$Changerate;
 echo "<br>job tax: ".$taxIncome;
 echo "<br>server name: ".$serverName;
 echo "<br>link to market: ".$linktopageJob;
 echo "<br>market country ID: ".$marketCountryID;
 echo "<br>market country Name: ".$countryName;
 echo "<br>time Job: ".$timeJob;
 echo "<br>day time Job: ".$dayTimeJob;
 echo "<br>gold Job: ".$goldJob;
 echo "<br>net gold job: ".$netgoldJob;
 echo "<br>link to employer: ".$linktoEmployer;
 echo "<br>currency Sell: ".$Currencysell;
 echo "<br>**********************END POST**********************<br><br>";
 echo_flush();

 echo "check $serverName";
 if ($stmt = $conn->prepare($sqlfind)) {
      $stmt->bind_param('sississiiiiiiis', $iscountryID, $countryName, $countryCurrency, $empCompanyID, $empCompanyName, $empProduct, $empQuality, $empSkill, $empSalary, $Changerate, $goldJob, $taxIncome, $netgoldJob, $serverName, $dayTimeJob);

     /* execute query */
     $stmt->execute();

     /* Store the result (to get properties) */
     $stmt->store_result();

     /* Get the number of rows */
     $num_of_rows = $stmt->num_rows;

     echo "<br>there are $num_of_rows results corresponding to requested search<br>";

     if ($num_of_rows == 0) {
        echo "THERE AREN'T RESULTS!";

        // SQLI INSERT
        $sqlJobIns = "INSERT INTO cJM".$serverName." (iscountryID, countryName, countryCurrency, empCompanyID, empCompanyName, empProduct, empQuality, empSkill, empSalary, goldExchange, salaryinGold, jobTax, netSalary, Server, link_to_page, link_to_employer, time_of_record, DateSave) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, CURDATE())";  // ADD NEW RECORDS

        if($stmt = $conn->prepare($sqlJobIns)) {
            $stmt->bind_param('ississiiiiiiiisss', $iscountryID, $countryName, $countryCurrency, $empCompanyID, $empCompanyName, $empProduct, $empQuality, $empSkill, $empSalary, $Changerate, $goldJob, $taxIncome, $netgoldJob, $serverName, $linktopageJob, $linktoEmployer, $dayTimeJob);

            /* execute query */
            $stmt->execute();
            $stmt->store_result();

            echo "<br>check $serverName";

            //var_dump($stmt);
            $num_of_rows_INS = $stmt->num_rows; 
            echo "<br>Check if data have been inserted: ";

            if ($stmt) {
                // RETURN AMOUNT OF DATA INSERT
                printf("added %d\n records to the DB...<br>", $stmt->affected_rows);        
            } else {
                echo "<br>there was a problem with data insert...<br>";
        }}
     } else echo ("<br>WE HAVE RESOULTS!");

     /* Bind the result to variables */
     $stmt->bind_result($idDB);

     while ($stmt->fetch()) {
        echo "<br>id from DATABASE is: ".$idDB;
     };
    /* free results */
    $stmt->free_result();
    /* close statement */
    $stmt->close();
 }};

感谢您的宝贵时间。希望此代码对其他人有所帮助。

阿尔伯托

【问题讨论】：

您的问题有点令人困惑，但让我们试一试。你能显示print_r($allJobs); 和print_r($result); 的结果吗？并尝试画出它“应该是”的样子。
哦，是的，谢谢。我会更新问题
可以用employee id作为数组键吗？比分配副本时，它只会重新分配密钥。使用array_vales 取回数组，但带有数字索引
不能这样做，因为我需要插入的所有值都是重复的，而不仅仅是公司 ID 或名称。有可能有前任。相同的名称，相同的 ID，相同的产品和技能，但不同的 SALARY，因为可能有两个相同的报价，仅针对提供给工人的 SALARY...
请从您的标题和问题中删除“已解决”。您可以自行发布答案，甚至接受它。对于有类似问题的其他人来说，这是正确的指示。

标签： php associative-array array-push

【解决方案1】：

如果产品、技能和薪水都与雇主提供的工作相关，您可以像这样填充您的数组：（注意它使用简写作为变量名）

在遍历 json 数组时使用它。

if (isset ($employers ['employer id']))
    $employers ['employer id'] = array ( 'name' => 'employer name', 'jobs' => array ('skill' => 'the skill', 'quality' => 'the quality', 'product' => 'the product', 'salary' => 'the salary'));
else
    $employers ['employer id']['jobs'][] = array('skill' => 'the skill', 'quality' => 'the quality', 'product' => 'the product', 'salary' => 'the salary'));

使用嵌套的 foreach 来获取数据。喜欢：

$allJobs = array ();
foreach ($employers as $employer) {
    for ($employer ['jobs'] as $job) {
        $allJobs[] = array(
            'EmployerName' => $employer['name'],
            'EmployerID' => $employer['id'],
            'Product' => $job['product'],
            'Quality' => $job['quality'],
            'Skill' => $job['skill'],
            'Salary' => $job['salary']);
    } 
}

【讨论】：

但是让 foreach 每次都创建具有不同名称的“最终数组”（如 $allJobs1、$allJobs2...）而不是将它们合并在一起不是更容易吗？
@alberto 你能解释一下你的意思吗？为什么需要 1 个以上的 allJobs 数组？
不，真的很抱歉我的糟糕解释。我的意思是：我需要获取要插入数据库的数据而无需重复。所以我想使用不同的数组来存储单个数据部分（例如雇主名称）。比组成一个数组检查“最终数组”中已有的内容。
问题是我需要重复这个操作 12 次，而不是有一个大数组，没有代表和覆盖其他循环数据，并使用它来获取所有要存储在数据库中的信息。这样更清晰一点？
像这样：1）第一个循环 -> 单个数组中的所有数据 -> 所有这些数组合并为一个，无需代表检查 in_array 是否存在相同的数据（薪水、产品、技能等级 ...）；比 2) 以技能 lvl 2... 比 3... 最多 12 级重新开始；比合并最后一个大数组中的所有内容并循环它以将数据插入到数据库中