Javascript 进入死循环

Question

当我取消选中输入复选框时，share.php javascript 进入无限循环打开新的 windows。

检查了所有 3 个都没有问题。

在某些时候出错了，但我找不到哪里。我不确定为什么会这样

<form action="functions/share.php" method="POST" target="_blank">
    <p>
        <textarea class="form-control" style="resize: none;height: 90px;padding: 10px;border-radius: 3px;" name="summary" required>Build a fully customizable Personal Website in seconds - for free.</textarea>
    </p>

    <div class="text-center">
        <input type="checkbox" name="LinkedIn" value="yes" checked><i class="fa fa-linkedin"></i> 
        <input type="checkbox" name="FaceBook" value="yes" checked><i class="fa fa-facebook"></i>
        <input type="checkbox" name="Twitter" value="yes" checked><i class="fa fa-twitter"></i>
    </div>

    <p><input type="submit" class="btn btn-primary btn-lg btn-block" value="Share" ></p>
</form>

share.php

<?php

    if(isset($_POST['summary'])){
        $linkedIn = "";
        $faceBook = "";
        $twitter = "";

        if($_POST['LinkedIn'] == "yes"){
            $linkedIn = "http://www.linkedin.com/";
        }

        if($_POST['FaceBook'] == "yes"){
            $faceBook = "http://www.facebook.com/";
        }

        if($_POST['Twitter'] == "yes"){
            $twitter = "http://www.twitter.com/";
        }

    }

?>
<html>
<body>
    <div id="linkedin" style="display: none;">
        <?php 
            echo htmlspecialchars($linkedIn); 
        ?>
    </div>
    <div id="facebook" style="display: none;">
        <?php 
            echo htmlspecialchars($faceBook); 
        ?>
    </div>
    <div id="twitter" style="display: none;">
        <?php 
            echo htmlspecialchars($twitter); 
        ?>
    </div>
    <script>

         var div = document.getElementById("linkedin");
         var linkedin= div.textContent;
         var div = document.getElementById("facebook");
         var faceBook= div.textContent;
         var div = document.getElementById("twitter");
         var twitter= div.textContent;

         if(linkedin != " "){

            window.open(linkedin);
         };
         if(faceBook != " "){

            window.open(faceBook);
         };
         if(twitter != " "){

            window.open(twitter);
         };

         close();

    </script>
</body>
</html>

Answer 1

首先，您必须将检查语句从

if($_POST['LinkedIn'] == "yes")

到 if(isset($_POST['LinkedIn']))

第二件事，你必须在你的JS代码中删除结束句close();

当您进行这两项更改时，选中的 link 将打开。如果有一个框未选中，则 window 将打开 link http://localhost/functions/share.php。要解决此问题，请在 'JS' 代码中声明 links。

您的最终代码：

<?php

    if(isset($_POST['summary'])){
        $linkedIn = "";
        $faceBook = "";
        $twitter = "";

        if(isset($_POST['LinkedIn']))
            $linkedIn = htmlspecialchars("http://www.linkedin.com/");
        else
            $linkedIn = "";

        if(isset($_POST['FaceBook']))
            $faceBook = htmlspecialchars("http://www.facebook.com/");
        else
            $faceBook = "";

        if(isset($_POST['Twitter']))
            $twitter = htmlspecialchars("http://www.twitter.com/");
        else
            $twitter = "";

    }

?>
<html>
<body>
    <script>

         var linkedin= "<?php echo $linkedIn; ?>";
         var faceBook= "<?php echo $faceBook; ?>";
         var twitter= "<?php echo $twitter; ?>";

         if(linkedin != "")
            window.open(linkedin);
         if(faceBook != "")
            window.open(faceBook);
         if(twitter != "")
            window.open(twitter);
         //close();

    </script>
</body>
</html>

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