Question

我使用servlet在谷歌浏览器中工作我有以下代码：

public void process(String input, PrintWriter out)
  {
        System.out.println(input.indexOf("\\\\"));
        String Json[]= input.split("\\\\");
        for(int x =0; x< Json.length;x++)
        {
            System.out.println(Json.length);
            //JSONProcess(json[x]);
            out.println("<p class=\"json\" style =\"display:none\"> ");
            out.println(Json[x]);
            out.println("</p>");
        }
        out.println("<script>"+
                "var JSONProcess= function(){\n" +
                "var JsonInfo = document.getElementByClassName(\"json\");\n" +
                "var canvasEl = document.getElementById(\"c\");\n" +
                "var nodesData =[];\n"+

                "window.alert(\"This is a test\");\n" +
                "console.log(JsonInfo.length);\n"+
                "for(var i = 0;i<JsonInfo.length; i++)\n" +
                "{\n" +
                "var Json = JsonInfo[x].innerHTML;\n" +
                "if(Json.source == null)\n" +
                "{\n" +
                "nodesData.push([Json.x, Json.y, Json.r, Json.id]);\n" +
                "}\n" +
                "}\n" +
                "for(var i=0; i<nodesData.length; i++)\n" +
                "{\n" +
                "   var data = nodesData[i];\n" +
                "   console.log(data[1]);\n" +
                "   console.log(data[2]);\n" +
                "   console.log(data[3]);\n" +
                "}\n" +
                "}</script>");
  }

虽然脚本标记不会引发任何错误，但是console.log或警报都不起作用。我试过了

删除window.log

这是作为对此问题的先前提问的答案，但没有做任何事情。控制台正在打印日志（我点击漏斗的东西来检查），我已关闭并重新打开铬。在这个问题的其他版本中提供的所有答案都没有奏效。

Answer 1

您必须调用该函数

 out.println("<script>"+
    "var JSONProcess= function(){\n" +
    "var JsonInfo = document.getElementByClassName(\"json\");\n" +
    "var canvasEl = document.getElementById(\"c\");\n" +
    "var nodesData =[];\n"+

    "window.alert(\"This is a test\");\n" +
    "console.log(JsonInfo.length);\n"+
    "for(var i = 0;i<JsonInfo.length; i++)\n" +
    "{\n" +
    "var Json = JsonInfo[x].innerHTML;\n" +
    "if(Json.source == null)\n" +
    "{\n" +
    "nodesData.push([Json.x, Json.y, Json.r, Json.id]);\n" +
    "}\n" +
    "}\n" +
    "for(var i=0; i<nodesData.length; i++)\n" +
    "{\n" +
    "   var data = nodesData[i];\n" +
    "   console.log(data[1]);\n" +
    "   console.log(data[2]);\n" +
    "   console.log(data[3]);\n" +
    "}\n" +
    "}" + 
    "JSONProcess()" + // here
    "</script>");

请注意，打印这样的javascript并不是一个好习惯，你最好将javascript放在.js文件中，而只打印带有源代码的单个脚本标记。

Console.log和Alert无法在Chrome中运行

1 个答案: