即使查询相同，PHP 返回的结果也少于 phpmyadmin

Question

当我在 phpmyadmin 中运行以下查询时，我得到了 18 个结果 - 所有这些都是正确的，也是我正在寻找的。但是，当我将查询复制并粘贴到一个 php 文件并运行该页面时，我得到了 17 个结果。

SELECT 
    ta.jobTaskID, 
    jt.customTaskTitle, jt.taskID, jt.status, 
    d.dealershipName, 
    j.jobNumber, j.jobID, j.title, j.jobSpec,
    wt.taskName, 
    po.dueToProduction
FROM taskassignments ta 
INNER JOIN jobTasks jt ON ta.jobTaskID = jt.jobTaskID
INNER JOIN jobs j ON jt.jobID = j.jobID
INNER JOIN dealerships d ON j.dealershipID = d.dealershipID
LEFT JOIN workflowtasks wt ON jt.taskID = wt.taskID
LEFT JOIN purchaseorders po ON j.jobID = po.jobID
WHERE ta.userID = 1 AND jt.status != 'Completed';

编辑:Here's a snapshot of my phpmyadmin result和 here's a snapshot of my var_dump .

这是我的 PHP 代码(我使用了我创建的数据库类)

$myTasks = $connection->runQuery("
SELECT 
    ta.jobTaskID, 
    jt.customTaskTitle, jt.taskID, jt.status, 
    d.dealershipName, 
    j.jobNumber, j.jobID, j.title, j.jobSpec,
    wt.taskName, 
    po.dueToProduction
FROM taskassignments ta 
INNER JOIN jobTasks jt ON ta.jobTaskID = jt.jobTaskID
INNER JOIN jobs j ON jt.jobID = j.jobID
INNER JOIN dealerships d ON j.dealershipID = d.dealershipID
LEFT JOIN workflowtasks wt ON jt.taskID = wt.taskID
LEFT JOIN purchaseorders po ON j.jobID = po.jobID
WHERE ta.userID = " . $userID . " AND jt.status != 'Completed'");

最后，这是连接类用来运行查询的代码:

// runs the user defined query
public function runQuery($runMe)
{
    $outArray = array();
    if ($this->checkConnection()) // if the connection is a resource
    {
        $returned = mysql_query($runMe, $this->dbConnection);
        if ($returned === false) // if there was an error during sql execution
        {
            echo "SQL Query error: " . mysql_error();
        }

        if ($returned === true) // if a update, insert, delete, etc... command was run
            return true;
        if (is_resource($returned)) 
        {
            $outArray = array(); // returned array with query results
            for ($i = 0; $i < mysql_num_rows($returned); $i++)
            {
                $outArray[] = mysql_fetch_assoc($returned);
            }
        }
        if (count($outArray) < 1)
            return null;
        else
            return $outArray;
    }
        else
            echo "Your Database Connection Was Unable To Be Authenticated.";
    }

Answer 1

更好的解决方案是根本不使用 mysql_num_rows() ，而只是不断重复调用 mysql_fetch_assoc() 直到它停止返回结果:

$outArray = array();
while ( $row = mysql_fetch_assoc($returned) ) {
    $outArray[] = $row;
}

(请注意，您的代码正在使用的 original PHP mysql API 已从 PHP 5.5.0 开始弃用，并将在未来的某个版本中删除。您真的应该移至 one of the supported APIs ，即 mysqli 或 PDO 。)