当我单击提交按钮时，我有表单，它会逐行显示所有数据 一切都工作正常，但我的问题是如何将所有值添加到 mysql 数据库 单击“提交到数据库”按钮。请任何人帮助我 这是我的完整代码
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script src="https://ajax.googleapis.com/ajax/libs/jqueryui/1.11.4/jquery-ui.min.js"></script>
<link rel="stylesheet" href="custom.css" type="text/css">
<html lang="en">
<head>
<meta charset="utf-8">
<title>Add Sales</title>
<script type="text/javascript">
function add_values(){
if(document.getElementById('edit_guid').value==""){
if( document.getElementById('stk').value!="" ){
if(document.getElementById('stk').value!=0){
sell=document.getElementById('pric').value;
disc=document.getElementById('stk').value;
item=document.getElementById('guid').value;
roll=parseInt(document.getElementById('roll_no').value);
$('<tr id='+item+'><td><lable id='+item+'roll >'+roll+'</label></td><td><input type=text readonly="readonly" value='+sell+'></td><td><input type=text readonly="readonly" value='+disc+' ></td></tr>').fadeIn("slow").appendTo('#item_copy_final');
document.getElementById('stk').value="";
document.getElementById('pric').value="";
document.getElementById('roll_no').value=roll+1;
document.getElementById('guid').value="";
}
}else{
alert('Please Select An Item');
}}}
</script>
<body>
<form name="form1" method="post" id="form1" action="">
<input type="hidden" id="roll_no" value="1" >
<div align="center">
<input type="hidden" id="guid">
<input type="hidden" id="edit_guid">
<table class="form" >
<tr>
<td>price</td>
<td>stk</td>
</tr>
<tr>
<td><input type='text' class='form-control' id="pric" name="pric"></td>
<td><input type='text' class='form-control' id="stk" name="stk"></td>
<td><input type="button" onclick="add_values()" id="add_new_code" value="submit" class="round"></div></form></td></tr>
</table>
<div style="overflow:auto ;max-height:300px; ">
<table class="form" id="item_copy_final" style="margin-left:45px "></table>
</div>
</div>
<div class="mytable_row ">
<form>
<div align="center">
<table>
<td><input type="button" value="submit_to_database" class="round"></td>
</table>
</div>
</form>
</body>
</html>最佳答案
更新您的代码 -
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script src="https://ajax.googleapis.com/ajax/libs/jqueryui/1.11.4/jquery-ui.min.js"></script>
<link rel="stylesheet" href="custom.css" type="text/css">
<html lang="en">
<head>
<meta charset="utf-8">
<title>Add Sales</title>
<script type="text/javascript">
function add_values(){
if(document.getElementById('edit_guid').value==""){
if( document.getElementById('stk').value!="" ){
if(document.getElementById('stk').value!=0){
sell=document.getElementById('pric').value;
disc=document.getElementById('stk').value;
item=document.getElementById('guid').value;
roll=parseInt(document.getElementById('roll_no').value);
$('<tr id='+item+'><td><lable id='+item+'roll >'+roll+'</label></td><td><input type=text name="price[]" readonly="readonly" value='+sell+'></td><td><input type=text name="stk[]" readonly="readonly" value='+disc+' ></td></tr>').fadeIn("slow").appendTo('#item_copy_final');
document.getElementById('stk').value="";
document.getElementById('pric').value="";
document.getElementById('roll_no').value=roll+1;
document.getElementById('guid').value="";
$("#pric").focus();
}
}
else{
alert('Please Select An Item');
}
}
}
</script>
</head>
<body>
<form name="form1" method="post" id="form1" action="">
<input type="hidden" id="roll_no" value="1" >
<div align="center">
<input type="hidden" id="guid">
<input type="hidden" id="edit_guid">
<table class="form" >
<tr><td>price</td><td>stk</td><td> </td></tr>
<tr>
<td><input type='text' class='form-control' id="pric" name="pric"></td>
<td><input type='text' class='form-control' id="stk" name="stk"></td>
<td><input type="button" onclick="add_values()" id="add_new_code" value="submit" class="round"></td>
</tr>
</table>
</div>
</form>
<div style="overflow:auto ;max-height:300px;" align="center">
<form method="post" action="test.php">
<table class="form" id="item_copy_final" style="margin-left:45px "></table>
<table>
<tr><td><input type="submit" name="submit_to_database" value="submit_to_database" class="round"></td></tr>
</table>
</form>
</div>
</body>
</html>
编写一个 PHP 脚本页面，在其中发布数据并将值添加到 mysql 数据库中。
“test.php”
<?php
mysql_connect("hostname", "database_user", "database_password") or die("Could not connect: " . mysql_error());
mysql_select_db("database_name");
if(isset($_POST["submit_to_database"])){
for($i=0; $i<count($_POST["price"]); $i++){
mysql_query("INSERT INTO `your_table` SET `your_price_field` = '".$_POST["price"][$i]."', `your_stk_field` = '".$_POST["stk"][$i]."'");
}
}
?>
关于javascript - 将数据插入mysql数据库，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/44092967/