blocks|key|2525073|text|php是在输出发送到浏览器之前/同时进行处理的，所以在构建页面的过程中删除了所有记录。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2525074|简而言之，您需要做的是(这只是解决它的一种方法...)：|2525075|2525076|为每个条目提供一个单独的表单|unordered-list-item|2525077|为条目的ID添加一个隐藏字段|2525078|将您的表单发布到另一个页面，该页面在用户提交表单|2525079|时执行删除操作|2525080|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Q|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|R|8|@]|9|@]|A|$]]|$1|E|3|F|5|G|7|S|8|@]|9|@]|A|$]]|$1|H|3|I|5|G|7|T|8|@]|9|@]|A|$]]|$1|J|3|K|5|G|7|U|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|V|8|@]|9|@]|A|$]]|$1|N|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|O|$]]

<p>The php is processed before / while the output is sent to the browser, so you are deleting all your records while your page is being built.</p>

<p>What you need to do in short is (this is just one way to solve it...):</p>

<ul>
<li>give each entry a separate form</li>
<li>add a hidden field for the ID of the entry</li>
<li>post your form to another page that does the deletion when the user submits the form</li>
</ul>


blocks|key|3360446|text|正如jeroen在评论中指出的那样，您正在混淆PHP和JS。您不能通过调用JS函数在页面上运行PHP，因为PHP是在页面显示之前解析的。您需要做的是使用onClick事件将数据回发到页面，以便它重新加载并调用代码来删除行：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3360447|<table+width="99%25"+border="0">
++<tr>
++++<td+align="center">ID</td>
++++<td+align="center">Username</td>
++++<td+align="center">EDIT</td>
++++<td+align="center">DELETE</td>
++</tr>

++<?+
++++if(isset($_POST['useId'])+&&+is_numeric($_POST['useId']))
++++{
+++++++mysql_query("DELETE+FROM+user+WHERE+useId+=+".+$_POST['useId']+."");
++++}

++++while+($fetch=mysql_fetch_array($q))
++++{
++?>++
++<form+method="post"+action="yourPHPScript.php"/>
++++<tr>
++++++<td+align="center"><?+echo+$fetch['useId'];+?></td>
++++++<td+align="center"><?+echo+$fetch['userUserName'];+?></td>
++++++<td+align="center"><?+echo+$fetch['useId'];+?></td>
++++++<input+type="hidden"+name="useId"+value="<?+echo+$fetch['useId'];+?>"/>
++++++<td+align="center">+<input+name="delete"+type="submit"+value="DELETE+"/></td>
++++</tr>
++</form>
++<?+}+?>

++</table>|code-block|syntax|javascript|3360448|我在这里所做的是修改您的代码，在页面加载时添加对POST变量useId的检查。一旦表单提交了页面，它就会存在。(我假设您在表单中使用method="POST“，如果您使用GET，则相应地更新)。请注意，应该在返回$q查询之前真正运行删除查询，以便在页面刷新后不会在列表中看到已删除的人员。|3360449|我还检查useId是否为数字(假设您的用户is是数字)作为安全检查，以确保不会有人向您的查询发送虚假数据。最后，我将隐藏字段添加到包含用户ID的表单中。|3360450|编辑:忘了声明表中的每一行都需要是单独的表单，这样才能正常工作，因为您只希望将该行的userID作为useID值发送。或者，您可以使用JS来确定要删除的行，并将其作为GET变量添加到userID上。|3360451|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

<p>As jeroen noted in the comments, you're mixing up your PHP and JS.  You cannot run PHP on a page by invoking a JS function since the PHP is parsed before the page is displayed.  What you need to do is use your onClick event to post data back to the page so that it reloads and calls the code to delete the row:</p>

<pre><code>&lt;table width="99%" border="0"&gt;
  &lt;tr&gt;
    &lt;td align="center"&gt;ID&lt;/td&gt;
    &lt;td align="center"&gt;Username&lt;/td&gt;
    &lt;td align="center"&gt;EDIT&lt;/td&gt;
    &lt;td align="center"&gt;DELETE&lt;/td&gt;
  &lt;/tr&gt;

  &lt;? 
    if(isset($_POST['useId']) &amp;&amp; is_numeric($_POST['useId']))
    {
       mysql_query("DELETE FROM user WHERE useId = ". $_POST['useId'] ."");
    }

    while ($fetch=mysql_fetch_array($q))
    {
  ?&gt;  
  &lt;form method="post" action="yourPHPScript.php"/&gt;
    &lt;tr&gt;
      &lt;td align="center"&gt;&lt;? echo $fetch['useId']; ?&gt;&lt;/td&gt;
      &lt;td align="center"&gt;&lt;? echo $fetch['userUserName']; ?&gt;&lt;/td&gt;
      &lt;td align="center"&gt;&lt;? echo $fetch['useId']; ?&gt;&lt;/td&gt;
      &lt;input type="hidden" name="useId" value="&lt;? echo $fetch['useId']; ?&gt;"/&gt;
      &lt;td align="center"&gt; &lt;input name="delete" type="submit" value="DELETE "/&gt;&lt;/td&gt;
    &lt;/tr&gt;
  &lt;/form&gt;
  &lt;? } ?&gt;

  &lt;/table&gt;
</code></pre>

<p>What I've done here is modified your code to add a check for the POST variable useId when the page loads.  This will exist once the page has been submitted by the form.  (I'm assuming you're using method="POST" in your form, if you're using GET then update accordingly).  Note that the delete query should really be run before you rurn your $q query so that you don't see the deleted people in your list after the page refreshes.</p>

<p>I'm also checking that useId is numeric (assuming your user IDs are numeric) as a security check to make sure someone doesn't send bogus data to your query.  Finally I added the hidden field to your form containing the user ID.</p>

<p>EDIT:  Forgot to also state that each row in your table needs to be a separate form for this to work properly since you want only the userID for that row to be send as the useID value.  Or alternatively you can use JS to determine which row you are deleting and tack on the userID as a GET var.</p>


<p>I am trying to make a PHP script that fetches records from the database, the fetching is working fine but the problem is that when I click <code>DELETE BUTTON</code>, it deletes all the records from the database, what I need to do here is to delete a specific record (the record on the row only), how can I do that?</p>

<p>Here is my code: </p>

<pre><code> &lt;table width="99%" border="0"&gt;
        &lt;tr&gt;
          &lt;td align="center"&gt;ID&lt;/td&gt;
          &lt;td align="center"&gt;Username&lt;/td&gt;
          &lt;td align="center"&gt;EDIT&lt;/td&gt;
          &lt;td align="center"&gt;DELETE&lt;/td&gt;
        &lt;/tr&gt;

          &lt;? 
      while ($fetch=mysql_fetch_array($q))
{
    ?&gt;          
        &lt;tr&gt;
          &lt;td align="center"&gt;&lt;? echo $fetch['useId']; ?&gt;&lt;/td&gt;
          &lt;td align="center"&gt;&lt;? echo $fetch['userUserName']; ?&gt;&lt;/td&gt;
          &lt;td align="center"&gt;&lt;? echo $fetch['useId']; ?&gt;&lt;/td&gt;
          &lt;td align="center"&gt; &lt;input name="delete" type="submit" value="DELETE " onclick="&lt;? mysql_query("DELETE FROM user WHERE useId = ". $fetch['useId'] ."") ?&gt; "/&gt;&lt;/td&gt;
          &lt;/tr&gt;
          &lt;? } ?&gt;

      &lt;/table&gt;
</code></pre>


Delete one record from MySQL Database

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在尝试制作一个从数据库中获取记录的PHP脚本，提取工作正常，但问题是，当我单击DELETE BUTTON时，它会删除数据库中的所有记录，我需要在这里做的是删除特定的记录(仅在行上的记录)，我该如何做？下面是我的代码： <table width="99%" border="0">        <tr>       ...

问从MySQL数据库中删除一条记录
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