Question

我的第一页删除用户query.php选择的查询，该网页完全正常：

<form method=post  action="delete.php"> 
List of queries<br/>
<?php
$ebits = ini_get('error_reporting');
error_reporting($ebits ^ E_NOTICE);
mysql_connect("localhost","root","") or die(mysql_error());
mysql_select_db("testdb") or die(mysql_error());
echo "<br />";
$query = "select * from queries ";
$result =  mysql_query($query) or die(mysql_error()); 
$count=mysql_num_rows($result);
while($row = mysql_fetch_array($result))  
{ 
print "<input type='checkbox' name='Query[]' value=\"".$row['queryId']."\"> "; 
echo "     ". $row['name']."      ". $row["address"]."           ". $row["contactNo"]."   

      ". $row["query"];
echo "<br />"; 
}  
?>
<input type="submit" value="Delete" name="Delete"> 
<br/>  
</form>

我尝试使用第二页delete.php的以下代码，但似乎没有任何效果代码1：

<?php
if($_POST['Delete'])  
{  
if(count($_POST['checkbox']) > 0) {  
   foreach($_POST['checkbox'] as $checkbox)  
   {  
        $del_id=$checkbox;  
        $sql = "DELETE * FROM queries WHERE `queryId`= '$del_id'";  
        $result = mysql_query($sql);  
        mysql_error();  

   }  
   echo "Selected Rows deleted";  
} else {  
     $NEW="Nothing to Delete";  
}

}
?>

代码2：

<?php
if(($_POST['Delete']))
{
$count=array();
$count=$_POST['checkbox'];
for($i=0;$i<count($count);$i++){
        $del_id = $checkbox[$i];
        $sql = "DELETE FROM queries WHERE queryId='$del_id' ";
        $result = mysql_query($sql);
}
$NEW="Selected records Deleted";
}
var_dump($_POST['checkbox']);
var_dump($count);

?>

Answer 1

您的复选框名称为“查询”，但您将其作为$_POST['checkbox']进行访问。这应该是$_POST['Query']。

编辑检查更新后的代码：

if($_POST['Delete']) { 
  if(count($_POST['Query']) > 0) { 
    foreach($_POST['Query'] as $checkbox) { 
      $del_id=$checkbox; 
      $sql = "DELETE * FROM queries WHERE queryId= '$del_id'"; 
      $result = mysql_query($sql); 
      mysql_error(); 
    } 
    echo "Selected Rows deleted"; 
  } 
  else { 
    $NEW="Nothing to Delete"; 
  } 
}

而不是：

$del_id=$checkbox;

这样做：

// if queryId is numeric
$del_id=intval($checkbox);

这可以确保您使用的值是数字，而不是来自用户的潜在恶意输入。我假设queryId是数字。如果不是，那么你需要这样做：

// if queryId is not numeric:
$del_id = mysql_real_escape_string($checkbox);

您的DELETE语法不正确：

$sql = "DELETE * FROM queries WHERE queryId= '$del_id'";

您只想DELETE FROM。此外，如果queryId的值是数字，则不需要围绕它的引号：

$sql = "DELETE FROM `queries` WHERE `queryId` = $del_id";

最后，你的MySQL错误调用没有做任何有用的事情：

mysql_error();

以下是您应该如何执行此操作以及其余代码：

if($_POST['Delete']) { 
  if(count($_POST['Query']) > 0) { 
    foreach($_POST['Query'] as $checkbox) { 
      $del_id= intval($checkbox); 
      $sql = "DELETE FROM `queries` WHERE `queryId` = $del_id"; 
      $result = mysql_query($sql); 
      if(!$result) {
        echo "There was an error in the query: " . mysql_error();
      }
    } 
    echo "Selected Rows deleted"; 
  } 
  else { 
    $NEW="Nothing to Delete"; 
  } 
}

要删除查询！我已经尝试过已有的帖子。它没用。

1 个答案: