当我单击按钮时，我已经制作了客户详细信息表单，它将 Json 数据发送给客户。但我的代码没有将数据插入数据库。我是网络技术新手，请告诉我哪里错了。
我的脚本:
<script>
$(document).ready(function(){
$("#btnBooking").on("click", function(){
var uName = document.getElementById('userName').value;
var mailId = document.getElementById('addressemailId').value;
var mobNum = document.getElementById('userContactNumber').value;
$.ajax({
url:"http://192.168.1.11/customerhomes/customer.php",
type:"GET",
dataType:"json",
data:{type:"booking",Name:uName, Email:mailId, Mob_Num:mobNum},
//type: should be same in server code, otherwise code will not run
ContentType:"application/json",
success: function(response){
alert("13");
},
error: function(err){
alert(JSON.stringify(err));
}
})
});
});
</script>
html 表单
<body>
<div class="page-header text-center">
<form >
<div class="col-lg-8">
<div class="form-group">
<label class="col-lg-3 control-label">Name:<font style="color: red;">*</font></label>
<div class="col-lg-9">
<input type="text" class="form-control" id="userName" name="userName" placeholder="Full Name" value="">
</div>
</div>
<div class="form-group">
<label class="col-lg-3 control-label">Mobile:<font style="color: red;">*</font></label>
<div class="col-lg-9">
<input type="text" class="form-control" id="userContactNumber" name="userContactNumber" type="number" placeholder="" onkeypress="enableKeys(event);" maxlength="10" placeholder="9966778888">
</div>
</div>
<div class="form-group">
<label class="col-lg-3 control-label">Email:<font style="color: red;">*</font></label>
<div class="col-lg-9">
<input type="text" class="form-control" name="addressemailId" id="addressemailId" placeholder="you@example.com" value="">
</div>
</div>
<div class="form-group marg-bot-45">
<label class="col-lg-3 control-label"></label>
<div class="col-lg-9">
<a href="" class="btn btn-info" id="btnBooking">Confirm Booking</a>
</div>
</div>
</div>
</form>
</div>
</body>
服务器代码
<?php
header('Access-Control-Allow-Origin: *');//Should work in Cross Domaim ajax Calling request
mysql_connect("localhost","root","1234");
mysql_select_db("customer_details");
if(isset($_GET['type']))
{
if($_GET['type']=="booking"){
$name = $_GET ['Name'];
$mail = $_GET ['Email'];
$mobile = $_GET ['Mob_Num'];
$query1 = "insert into customer(cust_name, cust_mobile, cust_email) values('$name','$mail','$mobile')";
$result1=mysql_query($query1);
}
}
else{
echo "Invalid format";
}
最佳答案
使用这个
JavaScript 代码:
<script>
$(document).ready(function(){
$("#btnBooking").on("click", function(e){
// as you have used hyperlink(a tag), this prevent to redirect to another/same page
e.preventDefault();
// get values from textboxs
var uName = $('#userName').val();
var mailId = $('#addressemailId').val();
var mobNum = $('#userContactNumber').val();
$.ajax({
url:"http://192.168.1.11/customerhomes/customer.php",
type:"GET",
dataType:"json",
data:{type:"booking",Name:uName, Email:mailId, Mob_Num:mobNum},
//type: should be same in server code, otherwise code will not run
ContentType:"application/json",
success: function(response){
alert(JSON.stringify(response));
},
error: function(err){
alert(JSON.stringify(err));
}
})
});
});
</script>
PHP 代码
<?php
header('Access-Control-Allow-Origin: *');//Should work in Cross Domaim ajax Calling request
mysql_connect("localhost","root","1234");
mysql_select_db("customer_details");
if(isset($_GET['type']))
{
$res = [];
if($_GET['type'] =="booking"){
$name = $_GET ['Name'];
$mail = $_GET ['Email'];
$mobile = $_GET ['Mob_Num'];
$query1 = "insert into customer(cust_name, cust_mobile, cust_email) values('$name','$mail','$mobile')";
$result1 = mysql_query($query1);
if($result1)
{
$res["flag"] = true;
$res["message"] = "Data Inserted Successfully";
}
else
{
$res["flag"] = false;
$res["message"] = "Oppes Errors";
}
}
}
else{
$res["flag"] = false;
$res["message"] = "Invalid format";
}
echo json_encode($res);
?>
如果数据插入成功，则返回 true 标志和消息，否则返回 false 标志和消息
关于javascript - 如何使用Jquery和Ajax将Json数据发送到数据库？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/27997839/