我在學習如何從引導模式提交表單數據到PHP文件。從我見過的其他問題來看,我認爲我是對的,但我一直在收到錯誤對話框。我必須失去一些明顯的東西。使用AJAX和JQuery發送模態表單內容到PHP文件時遇到困難
test.php的
<html>
<body>
<div class="container padding-top-10 change-width">
<div class="row padding-top-20" align="center">
<button class="btn btn-warning btn-lg" data-toggle="modal" data-target="#bandModal">Add Band(s)</button>
</div>
</div>
<div id="thanks"></div>
<div class="modal fade" id="bandModal" tabindex="-1" role="dialog" aria-labelledby="myModalLabel" aria-hidden="true">
<div class="modal-dialog modal-lg">
<div class="modal-content modal-lg">
<!-- Modal Header -->
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal">
<span aria-hidden="true">×</span>
<span class="sr-only">Close</span>
</button>
<h4 class="modal-title" id="bandModalLabel">
Add a Show
</h4>
</div>
<!-- Modal Body -->
<div class="modal-body row">
<div class="container col-md-12">
<form id="addBandForm">
<h3>Band Details<small>Enter each band name and primary contact information...</small></h3>
<div class="well" id="newBandRows">
<div class="row">
<div class="col-md-3">
<div class="form-group">
<label for "newBandName">Band Name:</label>
<input type="text" class="form-control" id="newBandName" name="newBandName" placeholder="Enter Band Name" />
</div>
</div>
<div class="col-md-3">
<div class="form-group">
<label for="primaryContact">Primary Contact:</label>
<input type="text" class="form-control" id="primaryContact" name="primaryContact" placeholder="Enter Name" />
</div>
</div>
<div class="col-md-3">
<div class="form-group">
<label for "personEmail">Primary Email:</label>
<input type="email" class="form-control" id="primaryEmail" name="primaryEmail" placeholder="Enter Email" />
</div>
</div>
<div class="col-md-3">
<div class="form-group">
<label for "personPhone">Primary Phone #:</label>
<input type="text" class="form-control" id="primaryPhone" name="primaryPhone" placeholder="Enter Phone #" />
</div>
</div>
</div>
</div>
<div id="newRowButton">
<div class="row">
<div class="col-md-1">
<button type="button" class="btn btn-success pull-left" onClick="addNewBandRow();">+</button>
</div>
<div id="remover" class="col-md-1">
</div>
<div class="col-md-7">
</div>
<div class="col-md-3 padding-top-10">
<button id="addBandSubmit" class="btn btn-primary pull-right">Submit</button>
</div>
</div>
</div>
<script src="js/newBand.js" type="text/javascript"></script>
</form>
</div>
</div>
<div class="modal-footer">
</div>
</div>
</div>
</div>
</body>
</html>
jQuery的
$(function() {
//twitter bootstrap script
$("#addBandSubmit").click(function() {
$.ajax({
type: "POST",
url: "womhScripts/addBand.php",
data: $('#addBandForm').serialize(),
success: function(msg) {
$("#thanks").html(msg)
$("#bandModal").modal('hide');
},
error: function(xhr, status, error) { var err = eval("(" + xhr.responseText + ")"); alert(err.Message); }
});
});
});
addBand.php
<?php
if (isset($_POST['newBandName'])) {
$bandName = strip_tags($_POST['newBandName']);
$contact = strip_tags($_POST['primaryContact']);
$email = strip_tags($_POST['primaryEmail']);
$phone = strip_tags($_POST['primaryPhone']);
echo "bandName =".$bandName."</br>";
echo "contact =".$contact."</br>";
echo "email =".$email."</br>";
echo "phone =".$phone."</br>";
echo "<span class="label label-info" >your message has been submitted .. Thanks you</span>";
}?>
看看你的開發者控制檯,看看裏面有沒有什麼 –
再看看這些'$ contact' - '$ primaryContact'和其他變量。錯誤報告應該拋出你未定義的變量通知,並因爲它而默默地失敗。 –
你得到了什麼錯誤?從你說的我理解的ajax請求返回一個error.Are你張貼在正確的路徑???你可以嘗試錯誤:function(xhr,status,error)var err = eval(「(」+ xhr.responseText + 「)」); alert(err.Message); }並告訴我們錯誤實際上是什麼。 –