我试图通过AJAX传递PHP回显值，但是在成功函数上，AJAX会跳过if和else条件并执行else语句，即使满足if条件也是如此。
下面是我的php和ajax代码
process_postadd.php
if(isset($_POST['btn-postAdd'])){
$istate = strip_tags($_POST['txt_istate']);
$iname = strip_tags($_POST['txt_iname']);
if(empty($iname)){
echo '1'; //check if iten name is empty
}
else{
try
{
if($insertItem = $postItem->postAdd($istate,$iname)){
echo "ok"; //sucessfully inserted
}
}
catch(PDOException $e)
{
echo $e->getMessage();
}
}
}
的script.js
$('document').ready(function()
{
/* validation */
$("#postAddForm").validate({
rules:
{
txt_istate: {
required: true
},
txt_iname: {
required: true
},
},
messages:
{
txt_istate: {
required: "state required"
},
txt_iname: {
required: "item name required"
},
},
submitHandler: submitForm
});
/* validation */
/* postadd submit */
function submitForm()
{
var data = $("#postAddForm").serialize();
$.ajax({
type : 'POST',
url : 'process_postadd.php',
data : data,
beforeSend: function()
{
$("#error").fadeOut();
$("#btn-postAdd").html('<span class="glyphicon glyphicon-transfer"></span> posting ...');
},
success : function(response)
{
if(response == "ok"){
$("#btn-postAdd").html('<img src="btn-ajax-loader.gif" /> Submiting item ...');
setTimeout(' window.location.href = "index.php"; ',4000);
}
else if(response == "1"){
$("#error").fadeIn(1000, function(){
$("#error").html('<div class="alert alert-danger"> <span class="glyphicon glyphicon-info-sign"></span> Sorry, add item name</div>');
$("#btn-postAdd").html('<span class="glyphicon glyphicon-log-in"></span> SIGN U');
});
}
else{
$("#error").fadeIn(1000, function(){
$("#error").html('<div class="alert alert-danger"> <span class="glyphicon glyphicon-info-sign"></span> '+response+' !</div>');
$("#btn-postAdd").html('<span class="glyphicon glyphicon-log-in"></span> Submit Item');
});
}
}
});
return false;
}
/* postadd submit */
});
HTML代码
<div class="col-sm-12">
<div id="error">
<!-- error will be shown here ! -->
</div>
<form method= "POST" class="form-horizontal" enctype="multipart/form-data" id="postAddForm">
<fieldset>
<div class="form-group">
<label class="col-md-3 control-label">Item condition</label>
<div class="col-md-8">
<label class="radio-inline" for="radios-00">
<input name="txt_istate" id="radios-00" value="New"
type="radio">
New</label>
<label class="radio-inline" for="radios-11">
<input name="txt_istate" id="radios-11" value="Used"
type="radio">
Used </label>
</div>
</div>
<div class="form-group">
<label class="col-md-3 control-label" for="Adtitle">Item Name</label>
<div class="col-md-8">
<input id="Adtitle" name="txt_iname" placeholder="name of item"
class="form-control input-md" type="text">
</div>
</div>
<div class="form-group">
<label class="col-md-3 control-label"></label>
<button class="btn btn-success" name="btn-postAdd" id="btn-postAdd" type="submit">
<span class="glyphicon glyphicon-log-in"></span> Submit Item
</button>
</div>
</fieldset>
</form>
</div>
答案 0 :(得分：0)
我不确定但是你的echo会尝试json_encode并添加一个console.log（响应）来检查返回
if($insertItem = $postItem->postAdd($istate,$iname)){
echo json_encode("ok"); //sucessfully inserted
}
和
if(empty($iname)){
echo json_encode('1'); //check if iten name is empty
}
答案 1 :(得分：0)
这应该有效，我不认为你的PHP实际上需要检查名称是否为空，你已经使用`jquery.validate（）;＆＃39;
关于您的成功，请删除此
else if(response == "1"){
$("#error").fadeIn(1000, function(){
$("#error").html('<div class="alert alert-danger"> <span class="glyphicon glyphicon-info-sign"></span> Sorry, add item name</div>');
$("#btn-postAdd").html('<span class="glyphicon glyphicon-log-in"></span> SIGN U');
});
}
以及你的php删除此
if(empty($iname)){
echo '1'; //check if iten name is empty
}
您已在validate()
然后您的代码将如下所示：
<script src="https://code.jquery.com/jquery-2.2.4.min.js" integrity="sha256-BbhdlvQf/xTY9gja0Dq3HiwQF8LaCRTXxZKRutelT44="
crossorigin="anonymous"></script>
<script type="text/javascript" src="admin_assets/js/validation.min.js"></script>// ignore this was for my test
<script type="text/javascript">
$('document').ready(function()
{
/* validation */
$("#postAddForm").validate({
rules:
{
txt_istate: {
required: true
},
txt_iname: {
required: true
},
},
messages:
{
txt_istate: {
required: "state required"
},
txt_iname: {
required: "item name required"
},
},
submitHandler: submitForm
});
/* validation */
/* postadd submit */
function submitForm()
{
var data = $("#postAddForm").serialize();
$.ajax({
type : 'POST',
url : 'process_postadd.php',
data : data,
beforeSend: function()
{
$("#error").fadeOut();
$("#btn-postAdd").html('<span class="glyphicon glyphicon-transfer"></span> posting ...');
},
success : function(response)
{
if(response == "ok"){
$("#btn-postAdd").html('<img src="btn-ajax-loader.gif" /> Submiting item ...');
setTimeout(' window.location.href = "index.php"; ',4000);
}
else{
$("#error").fadeIn(1000, function(){
$("#error").html('<div class="alert alert-danger"> <span class="glyphicon glyphicon-info-sign"></span> '+response+' !</div>');
$("#btn-postAdd").html('<span class="glyphicon glyphicon-log-in"></span> Submit Item');
});
}
}
});
return false;
}
/* postadd submit */
});
</script>
<div id="error"></div>
<form method="POST" id="postAddForm" action="">
<input type="text" name="txt_istate" id="txt_istate">
<input type="text" name="txt_iname" id="txt_iname">
<button type="submit" name="btn-postAdd" id="btn-postAdd">submit</button>
</form>
忽略我的表单，这是我的测试目的。您的表单已经
然后你的php
<?php
if (isset($_POST['btn-postAdd'])) {
$istate = strip_tags($_POST['txt_istate']);
$iname = strip_tags($_POST['txt_iname']);
try {
if ($insertItem = $postItem->postAdd($istate, $iname)) {
echo "ok"; //sucessfully inserted
} else {
echo "could not add item";
}
}
catch (PDOException $e) {
echo $e->getMessage();
}
}
?>
你现在好了。
另请考虑Auris的评论。