所以我做了简单的编辑/删除/新的mysql数据库更新。我正在使用ajax,因为我需要它在一个页面中。 因此,当我选中New按钮时,会出现一个新表单(从ajax调用到php文件)。我得到HTML格式。我想验证这一点。 但我不能。代码似乎是对的。我的代码在这里
trail.php
<?php include( 'connect.php'); ?>
<html>
<head>
<script src="http://code.jquery.com/jquery-latest.js"></script>
<script src="http://ajax.aspnetcdn.com/ajax/jquery.validate/1.11.1/jquery.validate.js"></script>
<script src="http://ajax.aspnetcdn.com/ajax/jquery.validate/1.11.1/jquery.validate.min.js"></script>
<script src="http://ajax.aspnetcdn.com/ajax/jquery.validate/1.11.1/additional-methods.js"></script>
<script src="http://ajax.aspnetcdn.com/ajax/jquery.validate/1.11.1/additional-methods.min.js"></script>
<script>
$(document).ready(function() {
$("#form1").validate({
debug: false,
rules: {
plid: "required",
},
messages: {
plid: "Please select a pack list id..",
},
submitHandler: function(form) {
$.ajax({
type: "POST",
url: "aa.php",
data: $('#form1').serialize(),
cache: false,
success: function(response) {
$('#result1').html(response);
}
});
}
});
});
</script>
</head>
<body>
<div id="result1"></div>Packing List</br>
<form id="form1" name="form1" action="" method="post">
<?php echo '<select name="plid" id="plid">'; echo '<option value="" selected="selected">--Select the Pack List Id--</option>'; $tempholder=a rray(); $sql="SELECT `pl_id`
FROM (
SELECT `pl_id`
FROM packlist
ORDER BY `pl_id` DESC
LIMIT 30
) AS t" ; $query=m ysql_query($sql) or die(mysql_error()); $nr=m ysql_num_rows($query); for ($i=0; $i<$nr; $i++){ $r=m ysql_fetch_array($query); if (!in_array($r[ 'pl_id'], $tempholder)){ $tempholder[$i]=$ r[ 'pl_id']; echo "<option>".$r[ "pl_id"]. "</option>"; } } echo '</select>'; ?>
<br/>
<input type="submit" name="new" id="new" value="New" />
<br/>
<input type="submit" name="delete" value="Delete" />
<br/>
<input type="submit" name="edit" id="edit" value="Edit" />
<br/>
</form>
</body>
我的ajax叫做php文件
<head>
<script src="http://code.jquery.com/jquery-latest.js"></script>
<script src="http://ajax.aspnetcdn.com/ajax/jquery.validate/1.11.1/jquery.validate.js"></script>
<script src="http://ajax.aspnetcdn.com/ajax/jquery.validate/1.11.1/jquery.validate.min.js"></script>
<script src="http://ajax.aspnetcdn.com/ajax/jquery.validate/1.11.1/additional-methods.js"></script>
<script src="http://ajax.aspnetcdn.com/ajax/jquery.validate/1.11.1/additional-methods.min.js"></script>
<script>
$(document).ready(function() {
$("#form2").validate({
debug: false,
rules: {
plidnew: "required",
},
messages: {
plidnew: "Please select a pack list id..",
}
});
});
</script>
</head>
<body>
<?php $a=isset($_POST[ 'plid']) && $_POST[ 'plid']; $b=isset($_POST[ 'new']) && $_POST[ 'new']; if($a&&$b) { ?>
<form name="form2" id="form2" method="post" action="">
<P>
<LABEL for="plidnew">PackList No
<INPUT type="text" id="plidnew">
</LABEL>
<BR>
<BR>
<LABEL for="itemidnew">Item Id
<INPUT type="text" id="itemidnew">
</LABEL>
<BR>
<BR>
<LABEL for="quannew">Quantity
<INPUT type="text" id="quannew">
</LABEL>
<BR>
<BR>
<LABEL for="potnew">Potency
<INPUT type="text" id="potnew">
</LABEL>
<BR>
<BR>
<LABEL for="sizenew">Size
<INPUT type="text" id="sizenew">
</LABEL>
<BR>
<BR>
<INPUT type="submit" id="newsubmit" name="newsubmit" value="Submit">
<INPUT type="reset">
</P>
</form>
<?php }
$c=isset($_POST[ 'plid']) && $_POST[ 'plid'];
$d=isset($_POST[ 'delete']) && $_POST[ 'delete'];
if($c&&$d) {
echo "delete!!";
}
$e=isset($_POST[ 'plid']) && $_POST[ 'plid'];
$f=isset($_POST[ 'edit']) && $_POST[ 'edit'];
if($e&&$f) {
?>
<form name="form3" id="form3" method="post" action="aa.php">
<P>
<LABEL for="plidedit">PackList No
<INPUT type="text" id="plidedit">
</LABEL>
<BR>
<BR>
<LABEL for="itemidedit">Item Id
<INPUT type="text" id="itemidedit">
</LABEL>
<BR>
<BR>
<LABEL for="quanedit">Quantity
<INPUT type="text" id="quanedit">
</LABEL>
<BR>
<BR>
<LABEL for="potedit">Potency
<INPUT type="text" id="potedit">
</LABEL>
<BR>
<BR>
<LABEL for="sizeedit">Size
<INPUT type="text" id="sizeedit">
</LABEL>
<BR>
<BR>
<INPUT type="submit" id="editsubmit" name="editsubmit" value="Submit">
<INPUT type="reset">
</P>
</form>
<?php } ?>
</body>
我对验证表格有疑问。我在两个页面都尝试了验证码。 但是没有像萤火虫那样的ajax效果。任何帮助表示感谢..非常感谢..
答案 0 :(得分:0)
将ajax
与type
“post”一起使用时,您应该小心传递数据
data
和data1
是web方法或服务的参数。试试这个
var d1="asd";
$.ajax({
type: "POST",
url: "aa.php",
data: "{data:'"+d1+"',data1:'"+d2+"'}",
cache: false,
success: function(response) {
$('#result1').html(response);
}
});