你能帮我解决这个问题吗?
假设我现在有这个:
HTML
<select name="insta_country_code" class="widget-user-country country-selector" data-account-name="
<?php echo $account1['Username'] ?>">
<option value="br">Brazil</option>
<option value="cl">Chile</option>
<option value="es" selected>Spain</option>
<option value="mx">Mexico</option>
<option value="gb">United Kingdom</option>
<option value="cr">Costa Rica</option>
</select>
JS
$(document).ready(function () {
$('.country-selector').change(function () {
var ig_name = $(this).data('account-name');
var country_code = $(this).val();
$.ajax({
type: 'POST',
dataType: 'json',
url: 'change-country.php',
data: {
insta_user: ig_name,
country_code: country_code,
},
success: function (data) {
if (data['status'] != 1) {
$.alert('server error');
}
},
error: function () {
$.alert('error');
}
});
})
});
我的.php
文件应该是什么?我需要这样做:
$InstagramCountry = $data['country_code'];
$InstagramUsername = $data['insta_user']);
$sql = mysql_query("UPDATE igaccounts SET Country='$InstagramCountry' WHERE Username='$InstagramUsername'");
非常感谢!
发布于 2019-03-04 13:48:14
试试这个,它应该对你的情况有效。我已经在我的本地主机上测试过了。对于您的HTML文件:
<select name="insta_country_code" class="widget-user-country country-selector" data-account-name="<?php echo $account1['Username'] ?>">
<option value="br">Brazil</option>
<option value="cl">Chile</option>
<option value="es" selected>Spain</option>
<option value="mx">Mexico</option>
<option value="gb">United Kingdom</option>
<option value="cr">Costa Rica</option>
</select>
对于您的JS文件:
<script>
$(document).ready(function () {
$('.country-selector').change(function () {
var ig_name = $(this).data('account-name');
var country_code = $(this).val();
$.ajax({
type: 'POST',
dataType: 'json',
url: 'change-country.php',
data: {
insta_user: ig_name,
country_code: country_code,
},
success: function (data) {
if (data['status'] != 1) {
$.alert('server error');
}
},
error: function () {
$.alert('error');
}
});
})
});
</script>
对于您的.PHP文件:
<?php
$data = $_REQUEST;
$InstagramCountry = $data['country_code'];
$InstagramUsername = $data['insta_user'];
echo "insta_user=".$InstagramUsername . " AND country_code=" . $InstagramCountry; exit;
如果还需要帮助就告诉我,兄弟。
发布于 2019-03-05 07:31:50
您好,您可以这样做:
你的php脚本:
if (isset($_POST["action"])) {
$action = $_POST["action"];
switch ($action) {
case 'SLC':
if (isset($_POST["id"])) {
$id = $_POST["id"];
if (is_int($id)) {
$query = "select * from alumni_users where userId = '$id' ";
$update = mysqli_query($mysqli, $query);
$response = array();
while($row = mysqli_fetch_array($update)){
.......
fill your response here
}
echo json_encode($response);
}
}
break;
}
}
其中action是要执行SLC、UPD、DEL等操作的命令,id是参数
然后在你的ajax中:
function getdetails() {
var value = $('#userId').val(); // value can be your array ! note: if you send a object json_encode(json_decode(,MyObj,true))
return $.ajax({
type: "POST",
url: "getInfo.php",
data: {action: "SLC",id: value }
})
}
这样叫它:
getdetails().done(function(response){
var data=JSON.parse(response);
if (data != null) {
//do somthing with your Data
}
})
希望能有所帮助
https://stackoverflow.com/questions/54975716
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