向 mysql 数据库输入动态字段失败，但我不明白为什么。 html 和 php 文件均已给出，请帮助我纠正代码并提供有关如何执行此操作的想法。
<html>
<head>
<title>Dynamically Add or Remove input fields in PHP with JQuery</title>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css" />
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/js/bootstrap.min.js"></script>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.2.0/jquery.min.js"></script>
</head>
<body>
<div class="container">
<br />
<br />
<h2 align="center">Purchase Entry</h2>
<div class="form-group">
<form name="add_name" id="add_name">
<div class="table-responsive">
<table class="table table-bordered" id="dynamic_field">
<tr>
<td><input type="text" name="name[]" placeholder="Invoice No." class="form-control name_list" /></td>
<td><input type="text" name="name[0]" placeholder="Description" class="form-control name_list" /></td>
<td><input type="text" name="name[1]" placeholder="Unit" class="form-control name_list" /></td>
<td><input type="text" name="name[2]" placeholder="Quantity" class="form-control name_list" /></td>
<td><input type="text" name="name[3]" placeholder="Amount" class="form-control name_list" /></td>
<td><button type="button" name="add" id="add" class="btn btn-success">Add More</button></td>
</tr>
</table>
<input type="button" name="submit" id="submit" class="btn btn-info" value="Add and Save" />
</div>
</form>
</div>
</div>
</body> </html> <script> $(document).ready(function(){
var i=1;
$('#add').click(function(){
i++;
$('#dynamic_field').append('<tr id="row'+i+'"><td><input type="text" name="name[]" placeholder="Invoice No." class="form-control name_list" /></td><td><input type="text" name="name[0]" placeholder="Description" class="form-control name_list" /></td><td><input type="text" name="name[1]" placeholder="Unit" class="form-control name_list" /></td></td><td><input type="text" name="name[2]" placeholder="Quantity" class="form-control name_list" /></td><td><input type="text" name="name[3]" placeholder="Amount" class="form-control name_list" /></td><td><button type="button" name="remove" id="'+i+'" class="btn btn-danger btn_remove">X</button></td></tr>');
});
$(document).on('click', '.btn_remove', function(){
var button_id = $(this).attr("id");
$('#row'+button_id+'').remove();
});
$('#submit').click(function(){
$.ajax({
url:"purchasesave.php",
method:"POST",
data:$('#add_name').serialize(),
success:function(data)
{
alert(data);
$('#add_name')[0].reset();
}
});
}); }); </script>
php 文件是这样的，但我想将每个字段添加到 mysql
<?php
include "includes/header.php";
include "includes/mydb.php";
include "includes/veriuser.php";
$conn = new mysqli($servername,$username,$password,$database)
?>
<?php
$number = count($_POST["purchase"]);
if($number > 0)
{
for($i=0; $i<$number; $i++)
{
if(trim($_POST["purchase"][$i] != ''))
{
$sql = "INSERT INTO tbl_name(purchase) VALUES('".mysqli_real_escape_string($conn, $_POST["name"][$i])."')";
mysqli_query($conn, $sql);
}
}
echo "Data Inserted";
}
else
{
echo "Please Enter Name";
}
?>
最佳答案
我发现一个问题需要更新
$.ajax({
url:"purchasesave.php",
method:"POST",
data:{ 'purchase' : $('#add_name').serialize() },
success:function(data)
{
alert(data);
$('#add_name')[0].reset();
}
});
这是常见格式，您使用 $number = count($_POST["purchase"]); 因此您需要在 ajax 请求中发送此变量。现在检查您的 ajax 请求
关于php - 向 mysql 数据库输入动态字段失败，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/45138240/