Question

我一直在搜索，我从堆栈溢出中发现了这个Question 所以我一直试图让它与我的代码一起工作，我似乎无法让它工作。

    html = '<li style="list-style:none;">';
                cmnt = this.comment.replace(new RegExp( "(" + preg_quote( firsttext ) + ")" , 'gi' ), "<span class='cutecl'>$1</span>");
                cmnt = cmnt.replace(new RegExp( "(" + preg_quote( secondtext ) + ")" , 'gi' ), "<span class='wincl'>$1</span>");
                cmnt = cmnt.replace(new RegExp( "(" + preg_quote( thirdtext ) + ")" , 'gi' ), "<span class='failcl'>$1</span>");
                html += cmnt;
                html += '<br/><a href="http://www.youtube.com/userPage.php?author='+escape(this.author)+'">'+this.author+'</a>';
                html += '<span class="label"> - '+(this.published.getMonth() + 1)+'/'+this.published.getDate()+'/'+this.published.getFullYear()+'</span>';
                html += '</li>';
                $('#comment').append(html);

当然：

function preg_quote( str ) {
return (str+'').replace(/([\\\.\+\*\?\[\^\]\$\(\)\{\}\=\!\<\>\|\:])/g, "\\$1");

}

当我有

时

 cmnt = this.comment.replace(new RegExp( "(" + preg_quote( firsttext ) + ")" , 'gi' ), "<span class='cutecl'>$1</span>");
            cmnt = cmnt.replace(new RegExp( "(" + preg_quote( secondtext ) + ")" , 'gi' ), "<span class='wincl'>$1</span>");
            cmnt = cmnt.replace(new RegExp( "(" + preg_quote( thirdtext ) + ")" , 'gi' ), "<span class='failcl'>$1</span>");

包含在我的代码中，当我想要它创建那个li时没有出现，但如果它消失了，它运行正常，任何线索我做错了什么？感谢提前，阿里

以下是问题的解答：http://jsfiddle.net/Yg8Qe/2/

Answer 1

从OP获得更多信息之后，这是修复：不要在替换字符串中对$1进行硬编码。

function preg_quote(str) {
    return (str + '').replace(/([\\\.\+\*\?\[\^\]\$\(\)\{\}\=\!\<\>\|\:])/g, "\\$1");
}

function wrap(data, search, before, after) {
    return data.replace( new RegExp( preg_quote( search ), 'gi' ), before + search + after );
}

var var1 = "var1",
    var2 = "var2",
    var3 = "var3",
    cmnt = "this comment will be replaced with var1 var2 var3";

cmnt = wrap(cmnt, 'var1', '<span class="cutecl">', '</span>');
cmnt = wrap(cmnt, 'var2', '<span class="wincl">', '</span>');
cmnt = wrap(cmnt, 'var3', '<span class="failcl">', '</span>');

$('#comment').append('<li style="list-style:none;">' + cmnt + '</li>');

这里有很多代码清理：缩进，分号，消除不必要的变量，并将“换行”逻辑放入一个漂亮的，可重用的函数中。

的 Demo → 的

~~看起来您在代码的第一行中遗漏了cmnt = this.comment.“无法正常工作。”~~

// ↓↓↓ right here cmnt = this.comment.replace(new RegExp( "(" + preg_quote( firsttext ) + ")" , 'gi' ), "$1"); cmnt = cmnt.replace(new RegExp( "(" + preg_quote( secondtext ) + ")" , 'gi' ), "$1"); cmnt = cmnt.replace(new RegExp( "(" + preg_quote( thirdtext ) + ")" , 'gi' ), "$1");

动态查找和替换文本Javascript（不区分大小写）

1 个答案: