计算两列并回显结果的脚本

Question

我有一个有效的 SQL 查询，我试图在一个小的 PHP 脚本中使用它，但遇到 Parse 错误，尝试了很多变体。希望你能帮忙。最终结果将是有一个包含“日期”和“通道号”的两个字段表单，然后给出给定日期的“通道”行数的结果计数。抱歉，相当新的 PHP/SQL，谢谢。

    <?php
// Connect to MSSQL and select the database
$link = mssql_connect('localhost', 'root', '', 'jm_db');
mssql_select_db('jm_db');

// Select all our records from a table

$mysql_query = mssql_query ('SELECT COUNT(*) FROM asterisk_cdr 
WHERE calldate LIKE '%2014-10-11%'
     AND channel LIKE '%SIP/4546975289%');

echo $sql;

?>

我已经重新编写了代码，但是得到“警告：mysql_fetch_array() 期望参数 1 是资源”和未定义的变量。

<?php

// Create connection
$mysqli = new mysqli($localhost, $root, $jm_db);

/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}

$sql = ("SELECT COUNT(*) FROM asterisk_cdr 
    WHERE calldate LIKE '%2014-10-11%'
    AND channel LIKE '%SIP/4546975289%'");

    $results= array();
while ($result = mysql_fetch_array($sql)) {
    $results[]= $result;
}

foreach($results as $result){
    echo $result['calldate'] . " " . $result['channel']; 
}

?>

Answer 1

您缺少一个引号（Stack 的语法突出显示向您显示），但它应该替换为一个开始的双引号并以相同的结尾。您不能使用所有单引号。

我用双引号替换了开始的单引号，以及匹配的结束双引号。

$mysql_query = mssql_query ("SELECT COUNT(*) FROM asterisk_cdr 
WHERE calldate LIKE '%2014-10-11%'
     AND channel LIKE '%SIP/4546975289%'");

作为旁注，您回显了错误的变量。

但是，这不是您回显结果的方式，而是使用循环。

类似的东西，并将Fieldname 替换为您要使用的：

while ($row = mssql_fetch_assoc($mysql_query)) {

    print $row['Fieldname'] . "\n";

}

或使用mssql_fetch_array()

你也可以使用：

$results= array();
while ($result = mssql_fetch_array($mysql_query)) {
    $results[]= $result;
}

foreach($results as $result){
    echo $result['calldate'] . " " . $result['channel']; 
}

---

有关Microsoft SQL Server功能的更多信息，请咨询：

• http://php.net/manual/en/book.mssql.php

Answer 2

    $mysql_query = mssql_query ('SELECT COUNT(*) FROM asterisk_cdr 
    WHERE calldate LIKE '%2014-10-11%'
    AND channel LIKE '%SIP/4546975289%');

    while($row=mssql_fetch_array($mysql_query))
    {
    echo $row[0];
    }

Answer 3

    $mysqli = new mysqli($servername, $username, $password, $dbname);

// Check connection
if ($mysqli->connect_error) {
    die("Connection failed: " . $mysqli->connect_error);
}

$mysql_query = mysqli ("SELECT COUNT(*) FROM asterisk_cdr WHERE calldate LIKE '%2014-10-11%' AND channel LIKE '%SIP/4546975289%'");

while ($row = mysql_fetch_array($mysql_query, MYSQL_ASSOC)) {
   echo ($row["channel"]);
}

Answer 4

这是一个使用 PDO 的简单示例

<?php
  try {
    $dns  = 'mysql:host=localhost;dbname=jm_db';
    $user = 'root';
    $pass = '';

    $options = array(
      PDO::MYSQL_ATTR_INIT_COMMAND => "SET NAMES utf8",
      PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION
    );
    $cnx = new PDO( $dns, $user, $pass, $options );
    $select = $cnx->query("SELECT COUNT(*) as count FROM asterisk_cdr WHERE calldate LIKE '%2014-10-11%' AND channel LIKE '%SIP/4546975289%'");

    $select->setFetchMode(PDO::FETCH_OBJ);

    while( $row = $select->fetch() )
    {
      echo '<h1>', $row->count , '</h1>';
    }
  } catch ( Exception $e ) {
    echo "Connect failed : ", $e->getMessage();
    die();
  }