JavaScript : Uncaught SyntaxError: Unexpected token <

Question

我有这些代码，它将 php 变量存储到全局 javascript 变量。所有模块都在同一个文件中，但我仍然在 rowNum1's 行收到 Uncaught SyntaxError: Unexpected token < 行:-

<script type="text/javascript">
    var glo = 0;
    var rowNum = <?php echo $_GET['SN'];?>;
    var newnames = new Array();
    var rowNum1 = "";
    var tempVal = "";
    var destTextarea = "";
    newnames=<?php echo json_encode($userinfoarr); ?>; //This varaible initialisation works fine


    function fetchNext(){
            if (glo++ === 0)
            {
                if(rowNum < newnames.length)
                {   
                    document.getElementById("Txt0").value = rowNum;
                    document.getElementById("Txt1").value = newnames[rowNum];
                    tempVal = newnames[rowNum];
                }
                rowNum++;
            }
            else
            {
                rowNum1=<?php echo json_encode($sln); ?>; //This one throws an Uncaught Syntax error
                if(rowNum1 < newnames.length)
                {   
                    document.getElementById("Txt0").value = rowNum1;
                    document.getElementById("Txt1").value = newnames[rowNum1];
                    tempVal = newnames[rowNum1];
                }
                rowNum1++;  
                console.log("test");
            }
    }
    </script>

这是 $sln 的 php 代码:

<?php
    if($_SERVER['REQUEST_METHOD'] == 'POST')
    {
        if(isset($_POST['Txt0']))
        {
            $sln = $_POST['Txt0'];
        }
    }
?>

这是 $userinfoarr 的 php 代码

<?php
    require_once('config.php');
    $result = mysqli_query($conn, "SELECT tweet from tweet");
    $userinfoarr = array();
    while ($row_user = mysqli_fetch_array($result,MYSQLI_ASSOC))
    {
        $userinfo = $row_user["tweet"];
        $push=array_push($userinfoarr, $userinfo);
    }
?>

有人可以帮我解决这个问题吗，因为我无法理解这个场景？这是浏览器内错误的屏幕截图

Answer 1

更改您的 php 代码如下。分配$sln默认值

<?php
    $sln = ""; //<---- assign default value to $sln
    if($_SERVER['REQUEST_METHOD'] == 'POST')
    {
        if(isset($_POST['Txt0']))
        {
            $sln = $_POST['Txt0'];
        }
    }
?>