即使查询相同，PHP返回的结果也比phpmyadmin少

Question

When I run the following query in phpmyadmin, I get back 18 results - all of which are correct and what I'm looking for. However, when I copy & paste the query into a php file and run the page, I get back 17 results.

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';

EDIT: Here's a snapshot of my phpmyadmin result and here's a snapshot of my var_dump .

Here's my PHP code (I use a DB class I created)

$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'");

Finally, here's the code the connection class uses to run a query:

// 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

A better solution is not to use mysql_num_rows() at all , but simply to keep calling mysql_fetch_assoc() repeatedly until it stops returning results:

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

(Ps. Just so you know, the original PHP mysql API , which your code is using, is deprecated as of PHP 5.5.0 and will be removed in some future version. You really should move to one of the supported APIs , either mysqli or PDO .)

Answer 2

Try replacing this line:

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

With:

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

Answer 3

Your last row is being skipped because you are only checking to see if $i is less than mysql_num_rows($returned) . Where you should be checking to see if it is less than or equal to.

So you will need to change the following line:

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

to

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

And then you will get all your data returned to your PHP as you would expect.

Answer 4

Try to clear all array used inside the loop by calling unset function after the loop to clear all array like

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

This will solve your problem as it solved mine.