0

我会马上说这个问题似乎解决了同样的情况,但对于我的生活,我没有看到解决方案/答案是什么。也许有人只需要为我更清楚地“拼出来”!

我有以下功能:

function CheckIfUrlExists(checkUrl) {
    var exists = '';
    $.ajax({
        type: 'POST',
        url: '../scripts/branchAdmin.php',
        data: {checkUrl: checkUrl},
        cache: false,
        success: function(response) {
            console.log('response: ' + response);
            if (response === 'true') {
                exists = 'true';
            } else if (response === 'false') {
                exists = 'false';
            }
        },
        error: function(response) {
            // return false and display error(s) from server
            exists = 'false';
        }
    });

    console.log('exists: ' + exists); // always displays empty string, regardless of what the php script returns

    return exists;
}

我用这个来称呼它:

var exists = CheckIfUrlExists($tr.find('input.editUrl').val());

if (exists === 'false') {
    //if (!CheckIfUrlExists($tr.find('input.editUrl').val())) {
    // New URL entered, verify user want to save it
    $('#confirmAddNewUrlDialog').data('row', $tr).dialog('open');
    ... // code to handle result
}

如何让“CheckIfUrlExist() 函数返回 true 或 false(或任何值),以便我可以在上面的代码中使用它?

4

1 回答 1

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将 ajax 调用异步选项设置为 false。然后你会得到结果。查看链接https://stackoverflow.com/a/1531710/399414

$.ajax({
        type: 'POST',
        async: false,
        url: '../scripts/branchAdmin.php',
...
于 2013-02-14T19:44:51.163 回答