Question

您好我正在尝试运行以下查询但似乎没有返回任何内容我想要的是从我的作业表中返回选择的job_type的job_discription。

任何帮助都会很棒，因为我花了好几个小时试图解决它。

谢谢

阿兰

<input type="hidden" name="JOB_TYPE" value="<?php print $_POST['JOB_TYPE'];?>"/>

<?php
$Query = " (SELECT JOB_TYPE, JOB_DISCRIPTION FROM jobs  " .
"WHERE jobs.JOB_TYPE ='$_POST[JOB_TYPE]' " .
"AND jobs.JOB_DISCRIPTION = 'JOB_DISCRIPTION')";

$Result = mysqli_query($DB, $Query);  
?>

<?php  
$Result = mysqli_query($DB,$Query)or die(mysqli_error($DB));

while ($Row = mysqli_fetch_assoc($Result))  // Now we go through the data displaying 
{

print  $Row ['JOB_DISCRIPTION']  ;  

}
?>

Answer 1

首先，代码很容易被sql注入：你不应该直接使用$ _POST数据。如果您想要特定类型的描述，请先删除最后一个条件。

Answer 2

从最后删除AND语句：

AND jobs.JOB_DISCRIPTION = 'JOB_DISCRIPTION'

同时从查询语句周围删除括号( )。

Answer 3

" -- quotation marks are only required at the start and end
SELECT JOB_TYPE
     , JOB_DISCRIPTION -- some people spell description with an 'e'
  FROM jobs  
 WHERE jobs.JOB_TYPE =$_POST['JOB_TYPE']    -- escape data (using modern methods) to prevent injection and note 
   AND jobs.JOB_DISCRIPTION = 'JOB_DISCRIPTION'; -- This is really strange
"

SQL查询没有返回任何内容

3 个答案: