Javascript AJAX调用不会传递成功或错误
我不明白为什么它不起作用。这让我发疯了 我有一个AJAX调用,它不会通过成功或错误Javascript AJAX调用不会传递成功或错误,javascript,jquery,ajax,Javascript,Jquery,Ajax,我不明白为什么它不起作用。这让我发疯了 我有一个AJAX调用,它不会通过成功或错误 function getStraatnamen(woonplaatsId) { var success = "not set"; $.ajax({ url: someURL, type: "POST", dataType: 'json', contentType: 'application/json', async:
function getStraatnamen(woonplaatsId) {
var success = "not set";
$.ajax({
url: someURL,
type: "POST",
dataType: 'json',
contentType: 'application/json',
async: false,
xhrFields: {
withCredentials: true
},
data: JSON.stringify({"woonplaatsId": woonplaatsId}),
succes: function (results) {
success = "true";
var straatnamen = results.GetStraatNamenResult;
var StraatNamenArray = [];
for (var i = 0; i < straatnamen.length; i++) {
var straatnaam = straatnamen[i];
StraatNamenArray.push({ text: straatnaam.naamField, value: straatnaam.idField });
}
console.log(StraatNamenArray);
straatnamenCombobox.setDataSource(StraatNamenArray);
straatnamenCombobox.bind("change", change_straatnaam);
},
error: function (error) {
success = "false";
console.log(error);
}
});
alert(success);
}
函数getStraatnamen(woonplaatsId){
var success=“未设置”;
$.ajax({
url:someURL,
类型:“POST”,
数据类型:“json”,
contentType:'应用程序/json',
async:false,
xhrFields:{
证书:正确
},
数据:JSON.stringify({“woonplaatsId”:woonplaatsId}),
成功:功能(结果){
success=“true”;
var straatnamen=results.GetStraatNamenResult;
var StraatNamenArray=[];
对于(变量i=0;ifunction getStraatnamen(woonplaatsId) {
var success = "not set";
$.ajax({
url: someURL,
type: "POST",
dataType: 'json',
contentType: 'application/json',
async: false,
xhrFields: {
withCredentials: true
},
data: JSON.stringify({"woonplaatsId": woonplaatsId}),
succes: function (results) {
success = "true";
var straatnamen = results.GetStraatNamenResult;
var StraatNamenArray = [];
for (var i = 0; i < straatnamen.length; i++) {
var straatnaam = straatnamen[i];
StraatNamenArray.push({ text: straatnaam.naamField, value: straatnaam.idField });
}
console.log(StraatNamenArray);
straatnamenCombobox.setDataSource(StraatNamenArray);
straatnamenCombobox.bind("change", change_straatnaam);
},
error: function (error) {
success = "false";
console.log(error);
}
});
alert(success);
}
有人知道为什么这件事很奇怪吗
Thanx.您在JQUERY中将“success”拼写为“succes”
换线
succes: function (results) {
到
你的脚本可能有用,但你打印错了。字段“成功”,结尾两个“S”。) 尝试“成功”而不是“成功”。愚蠢的错误,thanx