0
嗨,我想更新我的SQL表中的字段沒有在PHP頁面刷新我如何能實現我試過,但我的代碼不工作,我不知道我錯了,如何在php中刷新頁面刷新我的sql表?
的index.php
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<script src="http://code.jquery.com/jquery-1.9.1.js"></script>
<script>
$(document).ready(function(){
$('#myForm').on('submit',function(e) {
$.ajax({
url:'assignlead.php',
data:$(this).serialize(),
type:'POST',
success:function(data){
console.log(data);
$("#success").show().fadeOut(5000);
},
error:function(data){
$("#error").show().fadeOut(5000);
}
});
e.preventDefault();
});
});
</script>
</head>
<body>
<?php
include('conn.php');
$per_page = 3;
if($_GET)
{
$page=$_GET['page'];
}
$start = ($page-1)*$per_page;
$select_table = "select * from clientreg order by id limit $start,$per_page";
$variable = mysql_query($select_table);
?>
<form class="form2" action="" method="POST" name="myForm" id="myForm">
<div style="width:100%;">
<?php
$i=1;
while($row = mysql_fetch_array($variable))
{
?>
<input type="checkbox" name="users[]" value="<?php echo $row["id"]; ?>" >
<?php
}
?>
<div class="buttons"> <span id="error" style="display:none; color:#F00">Some Error!Please Fill form Properly </span> <span id="success" style="display:none; color:#0C0">All the records are submitted!</span>
<input class="greybutton" type="submit" value="Send" />
</div>
<?php
$sql = mysql_query("SELECT *FROM login where role=1");
while ($row = mysql_fetch_array($sql)){
?>
<input type="checkbox" name="eid[]" value="<?php echo $row["eid"]; ?>" ><?php echo $row["username"]; ?>
<?php
}
?>
</div>
</form>
</div>
</body>
</html>
assignlead.php
<?php
$conn = mysql_connect("localhost","root","root");
mysql_select_db("helixcrm",$conn);
if(isset($_POST["submit"]) && $_POST["submit"]!="") {
$usersCount = count($_POST["id"]);
for($i=0;$i<$usersCount;$i++) {
mysql_query("UPDATE clientreg set eid='" . $_POST["eid"][$i] . "' WHERE id='" . $_POST["id"][$i] . "'");
}
}
?>
<?php
$rowCount = count($_POST["users"]);
for($i=0;$i<$rowCount;$i++) {
$result = mysql_query("SELECT * FROM clientreg WHERE Id='" . $_POST["users"][$i] . "'");
$row[$i]= mysql_fetch_array($result);
$id=$row[$i]['id'];
?>
<input type="hidden" name="id[]" class="txtField" value="<?php echo $row[$i]['id']; ?>"></td>
<?php
$rowCoun = count($_POST["eid"]);
for($j=0;$j<$rowCoun;$j++) {
$result = mysql_query("SELECT * FROM login WHERE eid='" . $_POST["eid"][$j] . "'");
$row[$j]= mysql_fetch_array($result);
$eid=$row[$j]['eid'];
?>
<input type="hidden" name="eid[]" class="txtField" value="<?php echo $row[$j]['eid']; ?>">
<?php
}
}
?>
我嘗試了很多,但我不能讓我的輸出
我如何能實現我的輸出
在此先感謝
你必須使用AJAX。 – Jenz 2014-08-29 06:50:43
SQL注入發現...使用'prepared statements'與MySQLi或PDO; S – DanFromGermany 2014-08-29 07:23:43