持续警告：mysql_fetch_array（）预计参数1是资源

Question

我一直在试图找出是什么导致这个错误，但由于某种原因。它只是不会显示！

它已经给我这样的警告消息：

警告：mysql_fetch_array（）预计参数1是资源，布尔在C中给出：\ XAMPP \ htdocs中\ EIS \上线64 usercp.inc.php

我试过使用mysql_error();但它没有显示任何东西。

这里是我的代码：

<?php 
$firstname = getuserfield('txtFname'); 
$lastname = getuserfield('txtLname'); 
echo 'Hello '.$firstname.' '.$lastname.'.<br/>'; 

$anotif = 0; 
$dnotif = 0; 
$anotif = $anotif + getuserfield('approved_notif'); 
$dnotif = $dnotif + getuserfield('disapproved_notif'); 
?> 
<h3>Notifications:</h3> 
Approved:<?php if($anotif<1){echo 0;} else { echo '<a href="?anotif">'.$anotif.'</a>';}?><br/> 
<?php 
if(isset($_GET['anotif'])) { 
    $query = "SELECT * FROM hrf_leave WHERE empid = '$empid' AND formStatus = 1 AND checked = 0"; 
    $query_run = mysql_query($query) or die($query."<br/><br/>".mysql_error()); 
    echo "<table border=1> 
    <tr> 
    <th>Type of Leave</th> 
    <th>Specific Reason</th> 
    <th>Date From</th> 
    <th>Date To</th> 
    <th>Number of Days</th> 
    </tr>"; 
    echo "<tr>"; 
    while($record = mysql_fetch_array($query_run)){ // <------- this is line 64 
     $leave_id = $record['leave_id']; 
     echo "<td>" . $record['type_of_leave'] . "</td>"; 
     echo "<td>" . $record['specific_reason'] . "</td>"; 
     echo "<td>" . $record['date_from'] . "</td>"; 
     echo "<td>" . $record['date_to'] . "</td>"; 
     echo "<td>" . $record['num_of_days'] . "</td>"; 
     echo "</tr>"; 
     $query = "UPDATE hrf_leave SET checked=1 WHERE leave_id = $leave_id"; 
     if($query_run = mysql_query($query)){ 
      $anotif = $anotif-1; 
      $query = "UPDATE hrf_leave SET approved_notif=$anotif WHERE empid = $empid"; 
      if($query_run = mysql_query($query)){ 
       echo 'Recently approved requests'; 
      } 
     } 
    } 
    echo "</table>"; 
} 
?> 
Disapproved:<?php if($dnotif<1){echo 0;} else { echo '<a href="?dnotif">'.$dnotif.'</a><br/>';} ?> 
<?php 
if(isset($_GET['dnotif'])) { 
    $query = "SELECT * FROM hrf_leave WHERE empid = '$empid' AND formStatus = 2 AND checked=0"; 
    $query_run = mysql_query($query); 
    echo "<table border=1> 
    <tr> 
    <th>Type of Leave</th> 
    <th>Specific Reason</th> 
    <th>Date From</th> 
    <th>Date To</th> 
    <th>Number of Days</th> 
    </tr>"; 
    echo "<tr>"; 
    while($record = mysql_fetch_array($query_run)){ 
     $leave_id = $record['leave_id']; 
     echo "<td>" . $record['type_of_leave'] . "</td>"; 
     echo "<td>" . $record['specific_reason'] . "</td>"; 
     echo "<td>" . $record['date_from'] . "</td>"; 
     echo "<td>" . $record['date_to'] . "</td>"; 
     echo "<td>" . $record['num_of_days'] . "</td>"; 
     echo "</tr>"; 
     $query = "UPDATE hrf_leave SET checked=1 WHERE leave_id = $leave_id"; 
     if($query_run = mysql_query($query)){ 
      $dnotif = $dnotif-1; 
      $query = "UPDATE hrf_leave SET disapproved_notif=$dnotif WHERE empid = $empid"; 
      if($query_run = mysql_query($query)){ 
       echo 'Recently disapproved requests'; 
      } 
     } 
    } 
    echo "</table>"; 
} 
?> 

反正，我一直在试图输出的请假单卫生组织检查字段值= 0。一旦看出，系统将其设置为1。一旦链接被点击，所有批准和未批准的通知值都将等于零。

这是我的迷你通知系统。但我不能得到它，因为这个错误的工作..：/

Answer 1

在代码的followind块，变量$query_run，由产生该警告的代码行中，被覆盖：

$query = "UPDATE hrf_leave SET checked=1 WHERE leave_id = $leave_id"; 
    if($query_run = mysql_query($query)){ 
     $anotif = $anotif-1; 
     $query = "UPDATE hrf_leave SET approved_notif=$anotif WHERE empid = $empid"; 
     if($query_run = mysql_query($query)){ 
      echo 'Recently approved requests'; 
     } 
    } 

使用另一个变量名，因为$query_run已经由主要mysql_fetch_array通话

Answer 2

$query = "SELECT * FROM hrf_leave WHERE empid = '".$empid."' AND formStatus = 1 AND checked = 0"; 
echo $query; 

然后运行在你的phpmyadmin这个查询，你可以看到错误纠正。

您的查询很容易被sql注入。