Question

我正在尝试创建一个从mysql获取数据的php页面。我从数据库中获取对象列表，并以不同的形式显示每个对象以编辑值，然后仅保存编辑的值。但我无法提交表格。并且无法调用json部分和jaascript。

<!DOCTYPE html>
<html><head>
  <meta http-equiv="content-type" content="text/html; charset=UTF-8">
  <title>Expert Edit</title>


<script src="jquery-1.11.1.min.js"></script>
 <script>
  $("myforms").submit(function(){

    alert("before submit");  

    var data = {
      "tag": "editadsdata",
    };
    data = $(this).serialize() + "&" + $.param(data);

    $.ajax({
      type: "POST",
      dataType: "json",
      url: "http://localhost/expert.php", //Relative or absolute path to response.php file
      data: data,
      success: function(data) {

        alert("Form submitted successfully.");
      }
    });



            return false;
  });


</script>


</head>
<body>

<?php


$con=mysqli_connect("localhost","root","","expert");


        $result = mysqli_query($con,"select * from ads");
   $tmpCount = 1;
    while($row = mysqli_fetch_array($result)) {
  echo "<form action=\"\" class=\"myforms\" method=\"post\" accept-charset=\"utf-8\">";
  echo "<h2>New Entry ! $tmpcount</h2>";
  echo "URL : ";
  echo '<td><a href="' . $row['url'] . '">[Link]</a></td>';
  echo "<input hidden=\"true\" name=\"url\"".$row['url']."\">";
  echo "<br>";
  echo "IMAGE : ";
  echo "<img src=\"".$row['imagepath']."\" />";
  echo "<br>";
  echo "CATEGORY : ";
  echo "<input type=\"text\" name = \"category\" value=\"".$row['category']."\">";
  echo "<br>";
  echo "SUB-CATEGORY : ";
  echo "<input type=\"text\" name = \"subcategory\" value=\"".$row['subcategory']."\">";
  echo "<br>";
  echo "DESCRIPTION : ";
  echo "<input type=\"text\" name = \"description\" value=\"".$row['description']."\">";
  echo "<br>";
  echo "PRICE RANGE : ";
  echo $row['pricerange'];
  echo "<br>";
  echo "TAG : ";
  echo "<input type=\"text\" name = \"tags\" value=\"".$row['tag']."\">";
  echo "<br>";
  echo "UPLOADED BY : ";
  echo $row['uploadedby'];
  echo "<br>";
  echo "<input type=\"submit\" name=\"Submit\" value=\"submit\" />";  
  echo "</form>";
  $tmpcount++;
 }

 mysqli_close($con);
 ?>



  </body></html>

检查一下。这是有效的，但这是在HTML中。我想要它的PHP

<!DOCTYPE html>
<html>
<head>
 <meta http-equiv="content-type" content="text/html;            charset=UTF-8">
  <title>demo</title>

  <script type='text/javascript'    src='//code.jquery.com/jquery-1.10.1.js'></script>




  <link rel="stylesheet" type="text/css" href="/css/result-   light.css">

  <style type='text/css'>

  </style>



<script type='text/javascript'>//<![CDATA[ 
$(window).load(function(){
$('.myForms').submit(function () {
alert($(this).attr("id"));
return true;
});
});//]]>  

</script>


</head>
<body>
 <form action="http://www.google.com" method="get"    class="myForms" id="1stform">
    <input type="text" value="1st Form" name="q1" />
<input type="submit" value="Submit ALL" />
</form>
<form action="http://www.google.com" method="get"     class="myForms" id="2ndform">
 <input type="text" value="2nd Form" name="q2" />
<input type="submit" value="Submit ALL" />
</form>
<form action="http://www.google.com" method="get"  class="myForms" id="3rdform">
<input type="text" value="3rd Form" name="q3" />
<input type="submit" value="Submit ALL" />
</form>


</body>


</html>

Answer 1

因为您使用：$("myforms")。没有像这样的元素。如果myforms是您表单的ID，请使用：

$('#myforms')

如果这是它的类，那么

$('.myforms')

你在ajax中也有一些错误：

var data = {
  "tag": "editadsdata",
};
data = $(this).serialize() + "&" + $.param(data);

因为您定义的数据是json，并且您使用字符串覆盖数据，所以它将失败。

无法在php中提交表单

1 个答案: