我的搜索“功能”不起作用

My search "function" does not work

本文关键字：不起作用 功能 搜索 我的      更新时间：2023-09-26
 var cat = prompt("What Do You Want To Tweet??!");
 if(cat.length >= 140)
{
confirm("Sorry but your Tweet was over the 140 character limit by: " + (cat.length - 140));
confirm("You need to resubmit a Tweet that is not over the 140 Character limit");
}
else{
confirm("Success, Your Tweet has been posted!");
}
//scan for websites
if (cat.search(" www ", " http ", ".com"))
{
}
else{
confirm("We have Detected a URL in your Tweet");   
}

我的代码一直工作到我尝试"推特"一个网站。 如果我输入www或.com作为推文，它会告诉我它检测到了一个好的URL。但是当我输入 xxxxxx www xxxxxx(x 是其他任何东西(时，它不会检测到 URL。

你可以使用 jQuery 进行更准确的搜索。'.inArray((' 函数应该可以解决问题。

http://api.jquery.com/jquery.inarray/

<!doctype html>
<html lang="en">
<head>
    <meta charset="utf-8">
    <title>jQuery.inArray demo</title>
    <style>
        div {
            color: blue;
        }
        span {
            color: red;
        }
    </style>
    <script src="//code.jquery.com/jquery-1.10.2.js"></script>
</head>
<body>
    <div>"John" found at <span></span></div>
    <div>4 found at <span></span></div>
    <div>"Karl" not found, so <span></span></div>
    <div>"Pete" is in the array, but not at or after index 2, so <span></span></div>
    <script>
        var arr = [ 4, "Pete", 8, "John" ];
        var $spans = $( "span" );
        $spans.eq( 0 ).text( jQuery.inArray( "John", arr ) );
        $spans.eq( 1 ).text( jQuery.inArray( 4, arr ) );
        $spans.eq( 2 ).text( jQuery.inArray( "Karl", arr ) );
        $spans.eq( 3 ).text( jQuery.inArray( "Pete", arr, 2 ) );
    </script>
</body>
</html>

您可以尝试使用正则表达式

var cat = prompt("What Do You Want To Tweet??!"),
    expression = /[-a-zA-Z0-9@:%_'+.~#?&//=]{2,256}'.[a-z]{2,4}'b('/[-a-zA-Z0-9@:%_'+.~#?&//=]*)?/gi,
    regex = new RegExp(expression);
if(cat.length >= 140) {
    confirm("Sorry but your Tweet was over the 140 character limit by: " + (cat.length - 140));
    confirm("You need to resubmit a Tweet that is not over the 140 Character limit");
}
else{
    confirm("Success, Your Tweet has been posted!");
}
//scan for websites
if (cat.match(regex)) {
    confirm("We have dectected a URL in your Tweet");
}
else{
   confirm("no URL");   
}
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