PHP脚本未接收 – 或未读取 – POST JavaScript。

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英文:

PHP script doesn't receive - or doesn't read - POST JavaScript

问题

以下是翻译好的内容：

我试图将一个主体作为查询发送到一个PHP脚本，但当我将信息视为“$query['param']”时，此信息返回为未定义。

我用于发送信息的JavaScript代码如下：

  1. function insertNewRecord(data) {
  3.     fetch('/configs/database/add.php', {
  4.         method: 'POST',
  5.         headers: {
  6.           'Content-Type': 'application/x-www-form-urlencoded'
  7.         },
  8.         body: `name=${data.name}&contact=${data.contact}&delivery=${data.delivery}&value=${data.value}`
  9.     })
  10.     .then(function(response) {
  11.         if (response.ok) {
  12.             return response.text();
  13.         }
  14.         throw new Error('请求错误。');
  15.     })
  17.     var table = document.getElementById("employeeList").getElementsByTagName('tbody')[0];
  18.     var newRow = table.insertRow(table.length);
  19.     cell1 = newRow.insertCell(0);
  20.     cell1.innerHTML = data.name;
  21.     cell2 = newRow.insertCell(1);
  22.     cell2.innerHTML = data.contact;
  23.     cell3 = newRow.insertCell(2);
  24.     cell3.innerHTML = data.delivery;
  25.     cell4 = newRow.insertCell(3);
  26.     cell4.innerHTML = data.value;
  27.     cell4 = newRow.insertCell(4);
  28.     cell4.innerHTML = '<a onClick="onEdit(this)">编辑</a>' +
  29.                        '<a onClick="onDelete(this)">删除</a>';
  30. }

我用于接收请求的PHP脚本如下：

  1. <?php 
  3.     $protocol = ((!empty($_SERVER['HTTPS']) && $_SERVER['HTTPS'] != 'off') || $_SERVER['SERVER_PORT'] == 443) ? "https://" : "http://";  
  4.     $CurPageURL = $protocol . $_SERVER['HTTP_HOST'] . $_SERVER['REQUEST_URI'];  
  5.     $parts = parse_url($CurPageURL);
  7.     if(count($parts) <= 3) return;
  9.     parse_str($parts['query'], $query);
  11.     $conn = OpenCon();
  13.     $sql = "INSERT INTO users (name, contact, delivery, value) VALUES (" . $query['name'] . ", " . $query['contact'] . ", " . $query['delivery'] . ", " . $query['value'] . ")";
  14.     $result = $conn->query($sql);
  16. ?>

要修复这个问题，你可以尝试以下几个步骤：

  1. 确保你的JavaScript代码中的fetch请求成功发送到了PHP脚本。
  2. 在PHP脚本中，确保$query['name']、$query['contact']、$query['delivery']和$query['value']都包含在请求中，并且它们的值是正确的。
  3. 确保PHP脚本中的数据库连接（OpenCon()函数）正常工作，且没有错误。
  4. 如果问题仍然存在，可以在JavaScript代码中使用console.log输出来检查数据是否正确发送到PHP脚本，并在PHP脚本中使用var_dump或error_log来检查接收到的数据。

请确保你的代码中没有拼写错误，并根据错误消息来排除问题。

英文:

I'm trying to send a body as a query to a PHP script, but when I treat the information as "$query['param']", this information is returned as undefined.

The JavaScript code I'm using to send the information is as follows:

  1. function insertNewRecord(data) {
  3.     fetch(&#39;/configs/database/add.php&#39;, {
  4.         method: &#39;POST&#39;,
  5.         headers: {
  6.           &#39;Content-Type&#39;: &#39;application/x-www-form-urlencoded&#39;
  7.         },
  8.         body: `name=${data.name}&amp;contact=${data.contact}&amp;delivery=${data.delivery}&amp;value=${data.value}`
  9.     })
  10.     .then(function(response) {
  11.         if (response.ok) {
  12.             return response.text();
  13.         }
  14.         throw new Error(&#39;Error in the request.&#39;);
  15.     })
  17.     var table = document.getElementById(&quot;employeeList&quot;).getElementsByTagName(&#39;tbody&#39;)[0];
  18.     var newRow = table.insertRow(table.length);
  19.     cell1 = newRow.insertCell(0);
  20.     cell1.innerHTML = data.name;
  21.     cell2 = newRow.insertCell(1);
  22.     cell2.innerHTML = data.contact;
  23.     cell3 = newRow.insertCell(2);
  24.     cell3.innerHTML = data.delivery;
  25.     cell4 = newRow.insertCell(3);
  26.     cell4.innerHTML = data.value;
  27.     cell4 = newRow.insertCell(4);
  28.     cell4.innerHTML = `&lt;a onClick=&quot;onEdit(this)&quot;&gt;Edit&lt;/a&gt;
  29.                        &lt;a onClick=&quot;onDelete(this)&quot;&gt;Delete&lt;/a&gt;`;
  30. }

And the PHP script I'm using to receive the request is as follows:

  1. &lt;?php 
  3.     $protocol = ((!empty($_SERVER[&#39;HTTPS&#39;]) &amp;&amp; $_SERVER[&#39;HTTPS&#39;] != &#39;off&#39;) || $_SERVER[&#39;SERVER_PORT&#39;] == 443) ? &quot;https://&quot; : &quot;http://&quot;;  
  4.     $CurPageURL = $protocol . $_SERVER[&#39;HTTP_HOST&#39;] . $_SERVER[&#39;REQUEST_URI&#39;];  
  5.     $parts = parse_url($CurPageURL);
  7.     if(count($parts) &lt;= 3) return;
  9.     parse_str($parts[&#39;query&#39;], $query);
  11.     $conn = OpenCon();
  13.     $sql = &quot;INSERT INTO users (name, contact, delivery, value) VALUES (&quot; . $query[&#39;name&#39;] . &quot;, &quot; . $query[&#39;contact&#39;] . &quot;, &quot; . $query[&#39;delivery&#39;] . &quot;, &quot; . $query[&#39;value&#39;] . &quot;)&quot;;
  14.     $result = $conn-&gt;query($sql);
  16. ?&gt;

What should I do/change to fix this problem? I'm trying to do a CRUD in PHP, but this problem has been disturbing my work and I'm not able to solve it myself.

答案1

得分: 1

你要查找的数据在$_POST中，但你使用parse_str来从URL中读取值。这是代码中的问题。

在$_POST中检查POST数据，例如$_POST["name"]。
表单已经以POST方式提交，数据只会出现在URL中，如果你使用默认的GET方法来获取所有细节的URL。

英文:

The data you are looking for is in $_POST but you use parse_str to read the values out of the URL. That is the problem in the code.

Check for post data in $_POST. Like $_POST[&quot;name&quot;].
The form is POSTed and the data would only be in the URL if you did a fetch of a URL with all the details using the default GET method.

huangapple
  • 本文由 发表于 2023年6月22日 11:13:36
  • 转载请务必保留本文链接：https://go.coder-hub.com/76528384.html
  • fetch
  • forms
  • javascript
  • php
  • request
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