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javascript - Uncaught ReferenceError $ is not defined after new function

转载 作者:行者123 更新时间:2023-11-30 17:41:35 24 4
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我真的需要找到这个错误的根源。前几天我临时解决了它,但似乎每次我在 JS 文件中添加一个新函数时我都会得到它。之前我只会在新函数上得到“Uncaught ReferenceError $ is not defined”,但现在它适用于所有函数。我还收到“ncaught SyntaxError: Unexpected token <”,但至少目前这是次要问题。

索引文件

<html>
<head>
<meta content="text/html;charset=utf-8" http-equiv="Content-Type">
<meta content="utf-8" http-equiv="encoding">
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js"></script>
<script type="text/javascript" src="jquery-draw.js"></script>
</head>
<h2>Players</h2>
<div class="players">
<script>
setInterval( refreshPlayers, 5000 );
var inRequest = false;
</script></div>
<h3>Please select the Game you would like to view</h3>
<div class="pickGame">
<script>
setInterval( viewGames, 5000 );
var inRequestT = false;
</script>
</div>
<h2>Game</h2>
<div class="agonas">
<script>
setInterval( refreshAgonas, 5000);
var inRequestG = false;
</script></div>
</html>

jquery-draw.js文件

function viewGames(){
if ( inRequestT ) {
return false;
}
inRequestT = true;
var load = $.get('drawgames.php');
$(".pickGame").html('Refreshing');
load.error(function() {
console.log("Mlkia kaneis");
$(".pickGame").html('failed to load');
// do something here if request failed
});
load.success(function( res ) {
console.log( "Success" );
$(".pickGame").html('<form method="post" action="viewgame.php">
<select name="gameNo">'+res+'</select>
<input type="submit" value="Choose Game">
</form>');
});
load.done(function() {
console.log( "Completed" );
inRequestT = false;
});
}
function refreshAgonas() {
if ( inRequestG ) {
return false;
}
inRequestG = true;
var load = $.get('playersdata.php');
$(".agonas").html('Refreshing');
load.error(function() {
console.log("Mlkia kaneis");
$(".agonas").html('failed to load');
// do something here if request failed
});
load.success(function( res ) {
console.log( "Success" );
$(".agonas").html('<table border="1"><tr><th>ID</th><th>Name</th><th>Email</th><th>League</th><th>Sex</th><th>Birthday</th></tr>'+res+'</table>');
});
load.done(function() {
console.log( "Completed" );
inRequestG = false;
});
}
function refreshPlayers() {
if ( inRequest ) {
return false;
}
inRequest = true;
var load = $.get('playersdata.php');
$(".players").html('Refreshing');
load.error(function() {
console.log("Mlkia kaneis");
$(".players").html('failed to load');
// do something here if request failed
});
load.success(function( res ) {
console.log( "Success" );
$(".players").html('<table border="1"><tr><th>ID</th><th>Name</th><th>Email</th><th>League</th><th>Sex</th><th>Birthday</th></tr>'+res+'</table>');
});
load.done(function() {
console.log( "Completed" );
inRequest = false;
});
}

最佳答案

问题出在这里:

      $(".pickGame").html('<form method="post" action="viewgame.php">
<select name="gameNo">'+res+'</select>
<input type="submit" value="Choose Game">
</form>');

JavaScript 不允许在字符串中使用原始换行符,除非您使用反斜杠对它们进行转义。如果您有它们,则无法解析整个脚本,并且什么都不起作用。

单独地,您正在执行此操作:

var load = $.get('drawgames.php');

然后在 load 上调用函数 errorsuccessdoneload 将引用一个 jqXHR 对象。它们没有errorsuccess 函数(它们有done 函数)。您混淆了 $.ajax 函数接受回调的属性jqXHR 提供的函数。我推荐reviewing the documentation了解使用 $.get 的正确方法。

关于javascript - Uncaught ReferenceError $ is not defined after new function,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/20966768/

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