请参阅以下代码:
当我去编辑数据时,返回的保存数据是一个字段的状态'可以正确保存,或者有些字母在保存时会被删除。 在状态字段中,有三个可能的值' credit',' debit'或签证' '借记'和签证'值会被返回并正确保存,但保存的一些信用值就像' cr'即使文本框显示' credit'在保存之前...甚至清除字段并输入“信用卡”#39;在保存之前没有帮助..
CurrentPage.EditItem = function(id) {
if (confirm('Are you sure?')) {
console.log("DetailPage :: edit");
var name = $("#name").val();
var itemm = $("#itemm").val();
var amount = $("#amount").val();
var status = $("#status").val();
var Uid = localStorage.getItem("Uid");
//formData = {
// UId: sessionStorage.userId,
// itemm: $("#name").val(),
// amount: $("#amount").val(),
// status: $("#status").val(),
// }
if (itemm == "") {
alert("Please enter item");
} else if (amount == "") {
alert("Please enter office amount");
} else if (status == "") {
alert("Alert", "Please enter phone status");
} else {
$.ajax({
type:'POST',
url:'http://www.mywebsite.com/edit.php',
data:{'Uid':Uid,'itemm':itemm,'amount':amount,'status':status},
crossDomain: true, // enable this
success: function(data) {
alert("Edit item success");
window.location.href = "listof.html"
},
error: function() {
alert("Edit user gone wrong");
}
});
}
}
};
HTML
<div class="car">
<ul class="view">
<li >Item:</li>
<li ><input id="itemm" name="itemm" type="text"></li>
<li >Amount:</li>
<li ><input id="amount" name="amount" type="text"></li>
<li >Status:</li>
<li ><input id="status" name="status" type="text"></li>
</ul>
</div>
<div class="car" style="margin-bottom:40px;">
<button id="edit" class="btn btn-positive btn-block" onclick="currentPage.EditItem();">EDIT ITEM</button>
</div>
<div class="car" style="margin-bottom:40px;">
<button class="btn btn-negative btn-block" onclick="currentPage.deleteItem();">REMOVE ITEM</button>
PHP
<?php
header('Content-type: application/json');
$con=mysqli_connect("localhost","mxxxxxx","xxxxx89x78x","ccxcxcxcxc_db");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$itemm_id = $_POST['Uid'];
$itemm_itemm = $_POST['itemm'];
$itemm_amount = $_POST['amount'];
$itemm_status = $_POST['status'];
$qry = "UPDATE balance1 SET itemm = '$itemm_itemm', amount = '$itemm_amount' , status = '$itemm_status'
WHERE id ='$itemm_id'";
if (!mysqli_query($con,$sql))
{
die('Error: ' . mysqli_error($con));
}
echo "1 record updated";
mysqli_close($con);
?>
有什么想法???