將數據插入到mysql數據庫中
[英]insert data in to mysql database
問題描述
當我單擊提交按鈕時，我有表格，它逐行顯示所有工作正常的所有數據，但是我的問題是，當單擊“ submit_to_database”按鈕時，如何將所有這些值添加到mysql數據庫中。 請誰能幫我這是我的完整代碼
<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>
1 個解決方案
解決方案1
0 2017-05-21 04:23:30
更新您的代碼-
<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="";
$("#pric").focus();
}
}
else{
alert('Please Select An Item');
}
}
}
</script>
</head>
<body>
<form name="form1" method="post" id="form1" action="">
<input type="hidden" id="roll_no" value="1" >
<div align="center">
<input type="hidden" id="guid">
<input type="hidden" id="edit_guid">
<table class="form" >
<tr><td>price</td><td>stk</td><td> </td></tr>
<tr>
<td><input type='text' class='form-control' id="pric" name="pric"></td>
<td><input type='text' class='form-control' id="stk" name="stk"></td>
<td><input type="button" onclick="add_values()" id="add_new_code" value="submit" class="round"></td>
</tr>
</table>
</div>
</form>
<div style="overflow:auto ;max-height:300px;" align="center">
<form method="post" action="test.php">
<table class="form" id="item_copy_final" style="margin-left:45px "></table>
<table>
<tr><td><input type="submit" name="submit_to_database" value="submit_to_database" class="round"></td></tr>
</table>
</form>
</div>
</body>
</html>
編寫一個php腳本頁面，在其中發布數據並將值添加到mysql數據庫中。
“test.php的”
<?php
mysql_connect("hostname", "database_user", "database_password") or die("Could not connect: " . mysql_error());
mysql_select_db("database_name");
if(isset($_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]."'");
}
}
?>
在php中使用ajax向mysql數據庫插入數據
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