jQuery版本兼容性检测

是否有任何资源可以测试jQuery脚本/扩展/插件/版本兼容性问题？

刚刚写了一个小的jQuery插件来帮助解决版本兼容性问题...随意改进它.

(function($) {
    /**
     * Used for version test cases.
     * 
     * @param {string} left A string containing the version that will become
     *        the left hand operand.
     * @param {string} oper The comparison operator to test against. By
     *        default, the "==" operator will be used.
     * @param {string} right A string containing the version that will
     *        become the right hand operand. By default, the current jQuery
     *        version will be used.
     *        
     * @return {boolean} Returns the evaluation of the expression, either
     *         true or false.
     */
    $.isVersion = function(left, oper, right) {
        if (left) {
            var pre = /pre/i,
                replace = /[^\d]+/g,
                oper = oper || "==",
                right = right || $().jquery,
                l = left.replace(replace, ''),
                r = right.replace(replace, ''),
                l_len = l.length, r_len = r.length,
                l_pre = pre.test(left), r_pre = pre.test(right);

            l = (r_len > l_len ? parseInt(l) * ((r_len - l_len) * 10) : parseInt(l));
            r = (l_len > r_len ? parseInt(r) * ((l_len - r_len) * 10) : parseInt(r));

            switch(oper) {
                case "==": {
                    return (true === (l == r && (l_pre == r_pre)));
                }
                case ">=": {
                    return (true === (l >= r && (!l_pre || l_pre == r_pre)));
                }
                case "<=": {
                    return (true === (l <= r && (!r_pre || r_pre == l_pre)));
                }
                case ">": {
                    return (true === (l > r || (l == r && r_pre)));
                }
                case "<": {
                    return (true === (l < r || (l == r && l_pre)));
                }
            }
        }

        return false;
    }
})(jQuery);

可以像这样使用:

$.isVersion("1.4.2"); // returns true, if $().jquery == "1.4.2"
$.isVersion("1.3.2", ">"); // returns true if $().jquery > "1.3.2"
$.isVersion("1.3", ">", "1.2.6"); // returns true
$.isVersion("1.3.2", "<", "1.3.1"); // returns false
$.isVersion("1.4.0", ">=", "1.3.2"); // returns true
$.isVersion("1.4.1", "<=", "1.4.1"); // returns true

还支持预发布(版本比预发行版重,因此1.4.0pre <1.4.0):

$.isVersion("1.4.2", "<=", "1.4.2pre"); // returns false

代码有一些错误,特别是

$ .isVersion('1.9.2.17','<','2.0')返回false

$ .isVersion('1.17.2.1','>','1.8')返回false

解决方案
1.由附加的代码修复.
2.如果没有完整的重写,并不是那么棘手,无论如何都是一个罕见的案例.

(function($) {
/**
 * Used for version test cases.
 *
 * @param {string} left A string containing the version that will become
 *        the left hand operand.
 * @param {string} oper The comparison operator to test against. By
 *        default, the "==" operator will be used.
 * @param {string} right A string containing the version that will
 *        become the right hand operand. By default, the current jQuery
 *        version will be used.
 *
 * @return {boolean} Returns the evaluation of the expression, either
 *         true or false.
 */
$.isVersion = function(left, oper, right) {
    if (left) {
        var pre = /pre/i,
            replace = /[^\d]+/g,
            oper = oper || "==",
            right = right || $().jquery,
            l = left.replace(replace, ''),
            r = right.replace(replace, ''),
            l_len = l.length, r_len = r.length,
            l_pre = pre.test(left), r_pre = pre.test(right);

        l = (r_len > l_len ? parseInt(l) * Math.pow(10, (r_len - l_len)) : parseInt(l));
        r = (l_len > r_len ? parseInt(r) * Math.pow(10, (l_len - r_len)) : parseInt(r));

        switch(oper) {
            case "==": {
                return (true === (l == r && (l_pre == r_pre)));
            }
            case ">=": {
                return (true === (l >= r && (!l_pre || l_pre == r_pre)));
            }
            case "<=": {
                return (true === (l <= r && (!r_pre || r_pre == l_pre)));
            }
            case ">": {
                return (true === (l > r || (l == r && r_pre)));
            }
            case "<": {
                return (true === (l < r || (l == r && l_pre)));
            }
        }
    }

    return false;
}
})(jQuery);