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[英]MySQL returning Different results on PHP and PHPMyadmin using same Query
[英]PHP returning less results than phpmyadmin even though the query is the same
當我在phpmyadmin中運行以下查詢時,我得到18個結果-所有這些都是正確的以及我正在尋找的東西。 但是,當我將查詢復制並粘貼到php文件中並運行該頁面時,我得到了17個結果。
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';
編輯: 這是我的phpmyadmin結果 的快照,這是我的var_dump的快照 。
這是我的PHP代碼(我使用創建的DB類)
$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 = " . $userID . " AND jt.status != 'Completed'");
最后,這是連接類用於運行查詢的代碼:
// runs the user defined query
public function runQuery($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.";
}
更好的解決方案是不使用mysql_num_rows()
所有,而只是不斷的打電話mysql_fetch_assoc()
直到它停止返回結果:
$outArray = array();
while ( $row = mysql_fetch_assoc($returned) ) {
$outArray[] = $row;
}
(請注意。您知道,您的代碼正在使用的原始PHP mysql API從PHP 5.5.0開始不推薦使用,並將在將來的某些版本中刪除。您確實應該使用一種受支持的API ,即mysqli或PDO 。)
嘗試替換此行:
for ($i = 0; $i < mysql_num_rows($returned); $i++)
附:
for ($i = 0; $i < mysql_num_rows($returned) + 1; $i++)
由於僅檢查$i
是否小於mysql_num_rows($returned)
因此跳過了最后一行。 您應該在哪里查看它是否小於或等於。
因此,您需要更改以下行:
for ($i = 0; $i < mysql_num_rows($returned); $i++)
至
for ($i = 0; $i <= mysql_num_rows($returned); $i++)
然后,您將按照預期將所有數據返回到PHP。
嘗試清除循環中使用的所有數組,方法是在循環后調用unset函數以清除所有數組,例如
while(){
//your code
}
unset(array);
這將解決您的問題,因為它解決了我的問題。
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