PHP返回的结果比phpmyadmin少,即使查询是相同的

时间:2013-01-03 16:57:09

标签: php mysql phpmyadmin

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,我正在寻找什么。但是,当我复制&将查询粘贴到php文件并运行页面,我得到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 resulthere's a snapshot of my var_dump

这是我的PHP代码(我使用的是我创建的数据库类)

$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,这是连接类用于运行查询的代码:

// 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.";
    }

4 个答案:

答案 0 :(得分:1)

更好的解决方案是不要在所有中使用mysql_num_rows() ,而只是反复调用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。您知道,您的代码正在使用的original PHP mysql API自PHP 5.5.0起已弃用,将在以后的某个版本中删除。您真的应该转移到{{3} },one of the supported APIsmysqli。)

答案 1 :(得分:0)

尝试替换此行:

 for ($i = 0; $i < mysql_num_rows($returned); $i++)

使用:

 for ($i = 0; $i < mysql_num_rows($returned) + 1; $i++)

答案 2 :(得分:0)

您的最后一行被跳过,因为您只是检查$i是否小于mysql_num_rows($returned)。你应该检查它是否小于或等于。

因此您需要更改以下行:

for ($i = 0; $i < mysql_num_rows($returned); $i++)

for ($i = 0; $i <= mysql_num_rows($returned); $i++)

然后,您将按照预期将所有数据返回到PHP。

答案 3 :(得分:0)

尝试通过在循环后调用unset函数来清除循环内使用的所有数组,以清除所有数组,如

while(){
//your code
}
unset(array);

这将解决你的问题,因为它解决了我的问题。