HTML表单改变echo'ed变量，但不是数据库？

问题描述

 < div class =col-md-6> 
< h4>更新联络详情< / h4> 
< form action =method =postname =contactid =contact> 
< div class =form-group> 
< label> *地址：< / label> 
< input type =textclass =form-controlid =address_line_1name =address_line_1value =< ;? echo $ address_line_1;？> placeholder =地址行1 ...required =必需> 
< input type =textclass =form-controlid =address_line_2name =address_line_2value =< ;? echo $ address_line_2;？> placeholder =地址行2 ...> 
< label> *邮政编码：< / label> 
< input type =textclass =form-controlid =postcodename =postcodevalue =< ;? echo $ postcode;？> placeholder =邮编required =必填> 
< / div> 
< div class =form-group> 
< label> *电话号码：< / label> 
< input type =numberclass =form-controlid =contact_numbername =contact_numbervalue =< ;? echo $ contact_number;？> placeholder =电话号码required =必需> 
< / div> 
< div class =form-group> 
< input type =hiddenclass =form-controlid =useridname =useridvalue =< ;? echo $ userid;？>> 
< button type =submitid =contactname =contactclass =btn btn-primary btn-sm>保存详情< / button> 
< / div> 
< / div> 
  

运行至此脚本的文件：

<$ p $（$ _ POST ['contact']））{
$ address_line_1 = str_replace（'，\\\''，$ _ POST ['address_line_1' ]）;
$ address_line_2 = str_replace（'，\\'，$ _ POST ['address_line_2']）;
$ postcode = str_replace（'，\\'，$ _ POST ['postcode']）;
$ contact_number = $ _POST ['contact_number'];
$ uid = $ _POST ['userid'];

mysqli_query（$ conn，UPDATE ap_users SET address_line_1 ='$ address_line_1'，address_line_2 ='$ address_line_2'，postcode ='$ postcode'，$ contact_number ='$ contact_number'WHERE user_id ='$ UID'）;
}

非常基本，我知道 - 当我发送表单时，变量是在 echo $ address_line_1 这样的页面上回显，都会改变为新的结果，但是当我重新加载页面时，它们会返回到旧结果。看起来，当我发送表单时，MySQL数据库没有被更新，我不太清楚为什么？

解决方案

根据OP的要求

添加或 die（mysqli_error（$ conn））到 mysqli_query（）来查看是否有错误。

更新时也最好使用 affected_rows（）。

此外，您正在使用 if（isset（$ _ POST ['contact']））{并且具有表单和按钮的2个名称属性。

从< form> 删除一个。 < form> 上的名称属性仅适用于使用jQuery / Ajax的情况。

另外，如果您的表单和PHP / SQL位于同一个文件中，请使用标题重定向到同一页面，并确保您不是 * 在标题之前输出。并确保短标签已启用。

参考文献：

• *

---

您现在的代码可以打开。使用或，其中包含。

I have this form:

<div class="col-md-6">
                      <h4>Update Contact Details</h4>
                      <form action="" method="post" name="contact" id="contact">                      
<div class="form-group">
<label>*Address:</label>
<input type="text" class="form-control" id="address_line_1" name="address_line_1" value="<? echo $address_line_1; ?>" placeholder="Address Line 1..." required="required">
<input type="text" class="form-control" id="address_line_2" name="address_line_2" value="<? echo $address_line_2; ?>" placeholder="Address Line 2...">
<label>*Postcode:</label>
<input type="text" class="form-control" id="postcode" name="postcode" value="<? echo $postcode; ?>" placeholder="Postcode" required="required">
</div>
<div class="form-group">
<label>*Phone Number:</label>
<input type="number" class="form-control" id="contact_number" name="contact_number" value="<? echo $contact_number; ?>" placeholder="Phone Number" required="required">
</div>
<div class="form-group">
<input type="hidden" class="form-control" id="userid" name="userid" value="<? echo $userid; ?>">
<button type="submit" id="contact" name="contact" class="btn btn-primary btn-sm">Save Details</button>
</div>
                      </div>

Which runs to this script:

if(isset($_POST['contact'])){
    $address_line_1 = str_replace("'","\\'",$_POST['address_line_1']);
    $address_line_2 = str_replace("'","\\'",$_POST['address_line_2']);
    $postcode = str_replace("'","\\'",$_POST['postcode']);
    $contact_number = $_POST['contact_number'];
    $uid = $_POST['userid'];

    mysqli_query($conn, "UPDATE ap_users SET address_line_1 = '$address_line_1', address_line_2 = '$address_line_2', postcode = '$postcode', $contact_number = '$contact_number' WHERE user_id = '$uid'");
}

Very basic, I know - when I send the form, the variables which are echoed on the page such as echo $address_line_1 all change to the new results, although when I reload the page, they return to the old results. It appears that the MySQL database isn't being updated when I send the form and I'm not too sure why?

解决方案

As requested by the OP.

Add or die(mysqli_error($conn)) to mysqli_query() to see if errors come of it.

It's also best to use affected_rows() on update also.

Also, you are using if(isset($_POST['contact'])){ and have 2x name attributes for the form and the button.

Remove the one from <form>. Name attribute on <form> only works if using jQuery/Ajax.

Plus, if your form and PHP/SQL are in the same file, use a header to redirect to the same page and make sure you're not * outputting before header. and make sure short tags are enabled.

References:

• http://php.net/manual/en/mysqli.affected-rows.php
• http://php.net/manual/en/mysqli.error.php
• http://php.net/manual/en/function.header.php
• * How to fix "Headers already sent" error in PHP

---

Your present code is open to SQL injection. Use mysqli_* with prepared statements, or PDO with prepared statements.

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