Javascript Uncaught SyntaxError：意外的令牌

Question

我在使用此代碼時遇到問題，請告訴我有關 SyntaxError 的信息，但我在此代碼中沒有看到任何錯誤，並且在啟動 onchange 函數時無法使其正常工作

<script language="javascript" type="text/javascript">
        function checkSeLivre(){
            var ajaxRequest;  
            try{
                    ajaxRequest = new XMLHttpRequest();
            } catch (e){
                    try{
                            ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
                    } catch (e) {
                            try{
                                    ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
                            } catch (e){
                                    alert("Your browser broke!");
                                    return false;
                            }
                    }
            }
           ajaxRequest.onreadystatechange = function(){
                    if(ajaxRequest.readyState === 4 && ajaxRequest.status === 200){
                        var res = JSON.parse(ajaxRequest.responseText);
                        var result = res["status"];
                        var erro = res.message;
                        var ok = res.message;
                        console.log(result);
                        if(result == 'success'){
                            document.getElementById('ajaxDivOk').style.display = "block";
                            document.getElementById('ajaxDivOk').innerHTML = ok;
                            document.getElementById("ajaxDivErro").style.display = "none";

                        }
                        else{
                            document.getElementById('ajaxDivErro').style.display = "block";
                            document.getElementById('ajaxDivErro').innerHTML = erro;
                            document.getElementById("ajaxDivOk").style.display = "none";
                        }   
            }};
            var iduser = <?php echo $userid; ?>;
            var dia = document.getElementById('dia').value;
            var sala = document.getElementById('sala').value;
            var inicio = document.getElementById('inicio').value;
            var fim = document.getElementById('fim').value;
            var data = document.getElementById('data').value;    
            var queryString = "?iduser=" + iduser + "&dia=" + dia + "&sala=" + sala + "&inicio=" + inicio + "&fim=" + fim + "&data=" + data;
            ajaxRequest.open("GET", "checkSalaLivre.php" + queryString, true);
            ajaxRequest.send(null); 
        }
</script>

JavaScript 錯誤：Uncaught SyntaxError: Unexpected token < on line 37

第 37 行是

 var iduser = <?php echo $userid; ?>;

怎么了？

更新

javascript 和這個有什么區別？ 這個在另一個頁面中工作，但有不同的變量

 <script language="javascript" type="text/javascript">
        function checkSeLivre(){
            var ajaxRequest;  
            try{
                    ajaxRequest = new XMLHttpRequest();
            } catch (e){
                    try{
                            ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
                    } catch (e) {
                            try{
                                    ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
                            } catch (e){
                                    alert("Your browser broke!");
                                    return false;
                            }
                    }
            }
           ajaxRequest.onreadystatechange = function(){
                    if(ajaxRequest.readyState === 4 && ajaxRequest.status === 200){
                        var res = JSON.parse(ajaxRequest.responseText);
                        var result = res["status"];
                        var erro = res.message;
                        var ok = res.message;
                        console.log(result);
                        if(result === 'success'){
                            document.getElementById('ajaxDivOk').style.display = "block";
                            document.getElementById('ajaxDivOk').innerHTML = ok;
                            document.getElementById("ajaxDivErro").style.display = "none";

                        }
                        else{
                            document.getElementById('ajaxDivErro').style.display = "block";
                            document.getElementById('ajaxDivErro').innerHTML = erro;
                            document.getElementById("ajaxDivOk").style.display = "none";
                        }   
            }};
            var iduser = <?php echo $userid; ?>;
            var sala = document.getElementById('sala').value;
            var equip = document.getElementById('equip').value;
            var inicio = document.getElementById('inicio').value;
            var fim = document.getElementById('fim').value;
            var data = document.getElementById('data').value;
            var queryString = "?iduser=" + iduser + "&sala=" + sala  +"&equip=" + equip + "&inicio=" + inicio + "&fim=" + fim + "&data=" + data;
            ajaxRequest.open("GET", "checkEquipLivre.php" + queryString, true);
            ajaxRequest.send(null); 
        }
</script>

Answer 1

似乎您沒有通過 PHP 解釋器運行它。 違規行包含 PHP，並且不是有效的 JS。