代码崩溃 Firefox

Question

此脚本使 firefox 崩溃，并在 blueSelectorBar[mobIndex].style.border = "3px solid white"

的代码行显示无响应的脚本消息

这是脚本部分

        var blueSelectorBar = document.getElementById('blueSelector').getElementsByTagName('div');
        for (var bar = 0; bar < blueSelectorBar.length; bar++){
            var blueCurrentBox = blueSelectorBar[bar];
            console.log(blueCurrentBox.dataset.mob)
            var blueMobId = blueCurrentBox.dataset.mob
            if (eval(blueMobId) == toReturn){
                var mobIndex = bar
                for (var bar2 = 0; bar2 < blueSelectorBar.length; bar++){
                    blueSelectorBar[bar2].style.border = "3px solid black"
                };
            };
        };
        blueSelectorBar[mobIndex].style.border = "3px solid white"
        return toReturn

为什么会崩溃？

Answer 1

在内循环中：

            for (var bar2 = 0; bar2 < blueSelectorBar.length; bar++){
                blueSelectorBar[bar2].style.border = "3px solid black"
            };

您正在递增 bar 而不是 bar2。因此，该循环将永远不会结束，因为 bar2 永远不会改变。

代码崩溃 Firefox

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