php if 语句不适用于 jquery 变量

发布于 2024-11-16 21:38:52 字数 2025 浏览 0 评论 0原文

我发送到 PHP 的 jQuery 变量不起作用（或者至少，它似乎不起作用）。我已经用ajax将它发送到我的php。

请看一下，也许您可以看到问题：

                $('.do').click(function(){
                var cid2 = $(this).attr('id');
                var gebridauthpos = cid2.indexOf('||');
                var gebridauth = cid2.substring(gebridauthpos+2);
                $.post("agenda.php", {gebridauth: gebridauth});
                alert(gebridauth);
                <?php
                    if ($admin == true || isset($_POST['gebridauth']) AND $_SESSION['id'] == $_POST['gebridauth']) {
                        echo "$('#dialog').dialog('open');\n";
                        echo "var cid = $(this).attr('id');\n";
                        echo "var datum = cid.substr(0, 10);\n";
                        echo "var naampos = cid.indexOf('|');\n";
                        echo "var gebridpos = cid.indexOf('||');\n";
                        echo "var naam = cid.substring(naampos+1,gebridpos);\n";
                        echo "var gebrid = cid.substring(gebridpos+2);\n";
                        echo "$.ajax({\n";
                            echo "type: \"POST\",\n";
                            echo "url: \"agenda.php\",\n";
                            echo "data: naam,\n";
                            echo "success: function(){\n";
                                echo "$('#gebruikerinput').html(\"<input type='text' READONLY='' size='35' value='\" + naam +\"'>\");\n";
                                echo "$('#gebridinput').html(\"<input type='hidden' name='gebridtextbox' value='\" + gebrid + \"'>\");\n";
                                echo "$('#datuminput').html(\"<input type='text' READONLY='' size='12' name='datum' value='\" + datum + \"'>\");\n";
                            echo "}\n";
                        echo "})\n";
                        echo "return false;\n";
                    }
                ?>
            });

基本上我想做的是，当我单击 TD 时，在 PHP 的 if 语句中使用“gebridauth”。如果 TD 与登录者相同，则显示该对话框。

原文

The jQuery variable I send to my PHP doesn't work (Or atleast, it doesn't seem to work ). I've sent it to my php with ajax.

Please take a look at it, perhaps you can see the problem:

                $('.do').click(function(){
                var cid2 = $(this).attr('id');
                var gebridauthpos = cid2.indexOf('||');
                var gebridauth = cid2.substring(gebridauthpos+2);
                $.post("agenda.php", {gebridauth: gebridauth});
                alert(gebridauth);
                <?php
                    if ($admin == true || isset($_POST['gebridauth']) AND $_SESSION['id'] == $_POST['gebridauth']) {
                        echo "$('#dialog').dialog('open');\n";
                        echo "var cid = $(this).attr('id');\n";
                        echo "var datum = cid.substr(0, 10);\n";
                        echo "var naampos = cid.indexOf('|');\n";
                        echo "var gebridpos = cid.indexOf('||');\n";
                        echo "var naam = cid.substring(naampos+1,gebridpos);\n";
                        echo "var gebrid = cid.substring(gebridpos+2);\n";
                        echo "$.ajax({\n";
                            echo "type: \"POST\",\n";
                            echo "url: \"agenda.php\",\n";
                            echo "data: naam,\n";
                            echo "success: function(){\n";
                                echo "$('#gebruikerinput').html(\"<input type='text' READONLY='' size='35' value='\" + naam +\"'>\");\n";
                                echo "$('#gebridinput').html(\"<input type='hidden' name='gebridtextbox' value='\" + gebrid + \"'>\");\n";
                                echo "$('#datuminput').html(\"<input type='text' READONLY='' size='12' name='datum' value='\" + datum + \"'>\");\n";
                            echo "}\n";
                        echo "})\n";
                        echo "return false;\n";
                    }
                ?>
            });

Basically what I want to do, is using "gebridauth" in the if statement of my PHP when I click on a TD. If the TD is the same as the person that's logged in, show the dialog.

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

<逆流佳人身旁 2024-11-23 21:38:52

您需要对您的 $.post 调用进行回调，现在您只是发送 POST，而不关注服务器发回的内容，因此不会出现对话框。我认为你想要更多类似这样的东西（带有大注释的真实代码）：

$.post("agenda.php", {gebridauth: gebridauth}, function(data, textStatus, jqXHR) {
    // If the server sent back a "show the dialog" value in data then
    // show the dialog and all the other stuff that's currently in a
    // bunch of PHP echo calls.
});

You need a callback on your $.post call, right now you're just sending off the POST and not paying any attention to the what the server sends back so no dialog will appear. I think you want something more like this (with real code where the big comment is):

$.post("agenda.php", {gebridauth: gebridauth}, function(data, textStatus, jqXHR) {
    // If the server sent back a "show the dialog" value in data then
    // show the dialog and all the other stuff that's currently in a
    // bunch of PHP echo calls.
});

回复收藏 0 原文

羁绊已千年 2024-11-23 21:38:52

我认为您误解了 AJAX 的工作原理。你不能像这样混合 Javascript 和 PHP，因为它们在不同的系统上完全不同的时间运行。如果您要发布到 agenda.php，您的 PHP 代码需要位于文件 agenda.php 中。该文件不应包含 Javascript。您也无法像那样 echo Javascript 作为回报。