MySql Select 查询返回错误结果答案

【问题标题】：MySql Select query returning erroneous resultsMySql Select 查询返回错误结果
【发布时间】：2014-02-20 21:32:59
【问题描述】：

我创建了一个简单的 PHP 页面，它从 POST 请求中检索数据并将其保存在 MYSQL 数据库中。数据以正确的方式插入。现在，我想执行一个 Select 查询以从表中检索数据并将其显示在 UI 表中。 问题是 Select 语句出错。

这是错误：

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\abhideep_test_project\welcome.php on line 48

这是 PHP 代码：

<html>
<body>

<?php $name = $_POST["name"]; 
$email = $_POST["email"]; 
$mobile = $_POST["mobile"]; 
$address = $_POST["address"]; 
//=============Data Insertion=================

// Create connection
$con=mysqli_connect("localhost","root","","employee_db");

// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }
/*mysqli_query($con,"INSERT INTO employee_details (employee_name, employee_email, employee_mobile, employee_address) 
VALUES ('Glenn', 'glenn@gmail.com','9830098300','2/A, Work Lane')");*/

mysqli_query($con,"INSERT INTO employee_details (employee_name, employee_email, employee_mobile, employee_address) 
VALUES ('$name','$email','$mobile','$address')");

mysqli_close($con);
//=============Data Insertion=================


//=============Data Display=================
$con=mysqli_connect("localhost","root","","employee_db");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }


$result = mysqli_query($con,"SELECT  'employee_id' ,  'employee_name' ,  'employee_email' ,  'employee_mobile' ,  'employee_address' 
FROM  'employee_details' ");

echo "<table border='1'>
<tr>
<th>Name</th>
<th>Email</th>
<th>Mobile</th>
<th>Address</th>
</tr>";

while($row = mysqli_fetch_array($result))
  {
  echo "<tr>";
  echo "<td>" . $row['employee_name'] . "</td>";
  echo "<td>" . $row['employee_email'] . "</td>";
  echo "<td>" . $row['employee_mobile'] . "</td>";
  echo "<td>" . $row['employee_address'] . "</td>";
  echo "</tr>";
  }
echo "</table>";

mysqli_close($con);
//=============Data Display=================

?>
</body>
</html>

我哪里出错了？可以做些什么来获得所需的输出？

【问题讨论】：

标签： php mysql select

【解决方案1】：

您的选择查询在列和表名周围有单引号。您应该删除它们，或者用反引号替换它们：

"SELECT  `employee_id` ,  `employee_name` ,  `employee_email` ,  `employee_mobile` ,  'employee_address' 
FROM  `employee_details` "

更新：感谢@bartfriederichs 指出用单引号引用select 子句中的元素并非无效，因为查询将返回传递的相同数据。然而，用单引号引用表名是无效的。

我的回答假设您想要表employee_details 中的列employee_id, employee_name, employee_email, employee_mobile, employee_address 中的数据。

【讨论】：

周围的列不是无效的，它只会返回值。桌子周围是。
@BartFriederichs，是的，谢谢你看懂这句话。更新了答案，删除了不正确的陈述。

【解决方案2】：

这是不正确的

$result = mysqli_query($con,"SELECT  'employee_id' ,  'employee_name' ,  'employee_email' ,  'employee_mobile' ,  'employee_address' 
FROM  'employee_details' ");

应该是

$result = mysqli_query($con,"SELECT  `employee_id` ,  `employee_name` ,  `employee_email` ,  `employee_mobile` ,  `employee_address` 
FROM  `employee_details` ");

【讨论】：

【解决方案3】：

你的选择语句应该是这样的..使用backtick作为列名

$result = mysqli_query($con,"SELECT  `employee_id` ,  `employee_name` ,  `employee_email` ,  `employee_mobile` ,  `employee_address` 
FROM  `employee_details` ");

【讨论】：

【解决方案4】：

嘿，你需要像这样在插入变量中使用 addlashes 函数，

$name = addslashes($_POST["name"]); 
$email = addslashes($_POST["email"]); 
$mobile = addslashes($_POST["mobile"]); 
$address = addslashes($_POST["address"]);

然后运行您的查询，

mysqli_query($con,"INSERT INTO employee_details (employee_name, employee_email, employee_mobile, employee_address) 
VALUES ('$name','$email','$mobile','$address')");

【讨论】：

添加addlashes()的坏主意..为什么不使用mysqli_real_escape_string()
嘿，这不是个坏主意。我们应该在服务器端而不是在数据库中处理这些类型的事情。

【解决方案5】：

<html>
<body>
<?php 
$con = mysql_connect("localhost","root","") or die("mysql error");
        $db = mysql_select_db("employee_db",$con) or die ("problem in connecting database");
        ?>
<?php

$name = $_POST["name"]; 
$email = $_POST["email"]; 
$mobile = $_POST["mobile"]; 
$address = $_POST["address"]; 


$query =mysqli_query("INSERT INTO employee_details (employee_name, employee_email, employee_mobile, employee_address) 
VALUES ('$name','$email','$mobile','$address')");

?>
<table>
<thead>
<tr>
<th>Name</th>
<th>Email</th>
<th>Mobile</th>
<th>Address</th>
</tr>
</thead>

<tbody>

<?php
$sql ="SELECT  'employee_id' ,  'employee_name' ,  'employee_email' ,  'employee_mobile' 'employee_address' 
FROM  'employee_details' ";
                $runquery=mysql_query($sql);
                                    $rows= mysql_num_rows($runquery);

                                    if(is_resource($runquery) && $rows>0)       
                        {
                                    while($result= mysql_fetch_array($runquery))
                            {   


                                    echo ' <tr>';
                    echo '<td>'.$result['employee_name'].'</td>';
                    echo '<td>'. $result['employee_email'].'</td>';
                    echo '<td>'.$result['employee_mobile'].'</td>';
                    echo '<td>'.$result['employee_address'].'</td></tr>';

                            }}
            ?>             
                    </tbody>
                </table>

<?php mysqli_close($con);?>
<!-- =============Data Display=================-->

</body>
</html>

【讨论】：