jquery - jQuery版本兼容性检测

Question

是否有任何资源可以测试 jQuery 脚本/扩展/插件/任何版本兼容性问题？

Answer 1

刚刚写了一个小的 jQuery 插件来帮助解决版本兼容性问题......随时改进它。

(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（发布比预发布更重，因此 1.4.0pre < 1.4.0）：

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

Answer 2

The code has some bugs, notably

1. $.isVersion('1.9.2.17', '<', '2.0') returns false
2. $.isVersion('1.17.2.1', '>', '1.8') returns false

Solution
1. is fixed by the attached code.
2. is not as it's trickier without a complete re-write, and is a rarer case anyway.

(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);

Answer 3

至少我见过没有自动化工具可以做到这一点。这样做的原因是 jQuery 核心团队试图不引入重大更改，除非有真正的长期利益。这意味着当发生重大更改时，您想要的不是自动化系统总能告诉您的。

让我们以 jQuery 1.4 为例，这里有一个重大更改列表：http:
//jquery14.com/day-01/jquery-14#backwards

• jQuery()现在是空集，好还是坏？
• jQuery.browser.version现在是浏览器版本，好还是不好？
• JSON现在受到更严格的解析，好还是坏？

这些只是少数，但它们是否会破坏或帮助您的代码通常取决于。现在，如果你有.attr(val, func())那么显然这只是 1.4+，你可以检测到......所以一个可以确定你的代码可以使用的 jQuery 的最低版本的引擎可能是可能的。

检查兼容性问题，我认为这意味着在大多数情况下会破坏更改，这将更加困难，因为它们（大部分）本质上是非常奇怪或被破坏的极端情况......否则团队不会破坏它们： )

Answer 4

您是否想获取当前版本的 jQuery 并对其进行测试以查看其是否为某个版本？

$().jquery;

那将获得版本。

都会

Answer 5

这是我快速而肮脏的解决方案：

var versionGTE = function (valueA, valueB) {
  var values = [valueA, valueB];
  values.sort();
  // if valueA > valueB, values will have switched
  return (values[1] === valueA);
};

示例用法：

if (versionGTE(jQuery.fn.jquery, "1.3")) {
  // don't use @ in attr selectors
}

它对数组进行字母排序。唯一失败的情况是由于某种原因版本"x.y"vs "x.y.0". 在这种情况下，.0版本被认为更大。它也不支持"pre"版本。

这是一个较小的版本：

var versionGTE = function (valueA, valueB) {
  return ([valueA, valueB].sort()[1] === valueA);
};

如果您担心“rc1”、“pre”或x.0版本，这里有一个更可靠的函数：

var versionCompare = function (versionStringA, versionStringB) {
  // quick test of equality before digging in
  if (versionStringA === versionStringB) return 0;

  var versionedAlpha = /[a-z]+(\d+)/gi,
      getArray = function (verString) {
        // replace rc1, rc2, beta3, etc with .-1.1, .-1.2, .-1.3, etc
        return verString.replace(versionedAlpha, ".-1.$1").split(".");
      },
      valuesA = getArray(versionStringA),
      valuesB = getArray(versionStringB),
      maxLength = Math.max(valuesA.length, valuesB.length),
      hasLetters = /[a-z]/gi,
      // assume any version with letters is -1 (pre, rc, etc)
      // also assume that any null entries are 0 (1.5 === 1.5.0)
      parseVersion = function (verString) {
        return (verString) ? (hasLetters.test(verString)) ? -1 : parseInt(verString, 10) : 0;
      };

  // verify both arrays are the same size
  valuesA.length = maxLength;
  valuesB.length = maxLength;

  for (var i = 0; i < maxLength; i++) {
    var valueA = parseVersion(valuesA[i]),
        valueB = parseVersion(valuesB[i]);

    if (valueA < valueB) {
      return -1;
    } else if (valueA > valueB) {
      return 1;
    }
  }

  // all equal at this point
  return 0;
};

0这就像一个 sort 或 .compare 函数，如果相等，1如果 a > b 和-1如果 a < b ，它将返回。例子：

if (versionCompare(jQuery.fn.jquery, "1.3") >= 0) {
  // don't use @ in attr selectors
}

Answer 6

神圣的烟民，这些是有史以来最冗长的解决方案！我错过了什么吗？我必须。以下是我的解决方案，我错过了什么？

压缩版：

(parseInt(jQuery.fn.jquery.split('.').join('')) > 140) ? alert("Running jquery greater than 1.4.0") : alert("current jquery version is 1.4.0 or less");

为了清楚起见，长版本：

// get version as a string and get rid of the periods. 
version = jQuery.fn.jquery.split('.').join('');

// Make into one long number for easy comparison.  Example, 171, or 141.
version = parseInt(version);
if(version > 141){
    alert("We're using a version greater than 1.4.1");
}else{
    alert("jQuery version is 1.4.1 or lower");
}