Question

我是java和javascript的新手。我已经开始使用json对象从我们的SQL服务器返回数据，但它很好，但我遇到了一个问题，我不知道如何处理它。我甚至不确定它是否可能但是从来没有伤害过要求。

我在数据库中有一个存储HTML内容的字段。

<p>Please verify you have done the following before submitting a trouble-ticket</p>    <ol>   <li>Be sure you are using your correct user name and password (e.e., first initial and last name)</li>   <li>Password are case sensitive</li>   <li>Be sure to select the domain from the drop-down list.</li>   <li>Network cable must be plugged in!</li>  </ol>

我正在尝试将其输出到JSON对象，以便我可以在屏幕上显示它的当前格式。

我在.jsp页面中构建我的json对象，如下所示：

if(rs != null && rs.next()){

  out.print("{\"qteTips\" : [");
  tipList = new StringBuffer();
  int rowCount = 0;

  String HelpfulTipID = null;
  String Title = null;
  String Tip = null;
  String DateCreated = null;
  String Active = null;

  do{
    rowCount++;

    HelpfulTipID = rs.getString("HelpfulTipID");
    Title = rs.getString("Title");
    DateCreated = rs.getString("DateCreated");
    Active = rs.getString("Active");
    Tip = rs.getString("Tip");

    if(rowCount > 1){
      out.print(",");
    }

    out.print("{\"HelpfulTipID\":\""+HelpfulTipID+"\",");
    out.print("\"Title\":\""+Title+"\",");
    out.print("\"DateCreated\":\""+DateCreated+"\",");
    out.print("\"Active\":\""+Active+"\",");
    out.print("\"Tip\":\""+Tip+"\"}");
    if(DEBUG){
      System.out.println("created record #"+rowCount);
    }
  }
  while(rs != null && rs.next());
  out.print("]}");
  try   {rs.close();} catch (SQLException sqle){}
  try{ps.close();} catch (SQLException sqle){}  

}else{
  out.print("{\"error\":\"No tips found.\"}");
}

但当我抓住物体时，它看起来像这样：

{"qteTips" : [{"HelpfulTipID":"47","Title":"Unable to login to your comptuer?","DateCreated":"06/01/15","Active":"1","Tip":"<p>Please verify you have done the following before submitting a trouble-ticket</p>

<ol>
    <li>Be sure you are using your correct user name and password (e.e., first initial and last name)</li>
    <li>Password are case sensitive</li>
    <li>Be sure to select the domain from the drop-down list.</li>
    <li>Network cable must be plugged in!</li>
</ol>
"}]}

尝试验证时，我在jsonlint.com上收到此错误。

Parse error on line 8:
...            "Tip": "<p>Please verify yo
----------------------^
Expecting 'STRING', 'NUMBER', 'NULL', 'TRUE', 'FALSE', '{', '['

有没有人知道我想要做的更好的方法？

Answer 1

问题是返回的JSON中的换行符，您将不得不将它们转义为\n。我经常这样做：

line = line.replaceAll('\r', '').replaceAll('\n', '\\n')

基本上将\r字符替换为\n字符，\\n替换为...string goes here\nNext line，这将在您的JSON上呈现为Ti.API.debug('Installing sql database "' + dbFile + '" with name "' + dbName + '"'); var db = Ti.Database.install(dbFile, dbName); [DEBUG] Installing sql database "/trackit.sqlite" with name "trackit"

Answer 2

我已经包含了一个jsp页面。

<%@ page contentType="text/html"%>
<%@include file="GetAllHelpfulTips.jsp"%> 
<html lang="en">
<head>

该页面执行一个sql语句，该语句在循环时构建数据。

ps = conn.prepareStatement(sqlTask.toString());
			rs = ps.executeQuery();
			if(rs != null && rs.next()){

				tipList = new StringBuffer();
				int rowCount = 0;
				
				String HelpfulTipID = null;
				String Title = null;
				String Tip = null;
				String DateCreated = null;
				String Active = null;
				
				do{
					rowCount++;
							
					HelpfulTipID = rs.getString("HelpfulTipID");
					Title = rs.getString("Title");
					DateCreated = rs.getString("DateCreated");
					Active = rs.getString("Active");
					Tip = rs.getString("Tip");
						
					if(tipsPage == "1"){
						System.out.println("from tip page");
						tipList.append("<div id='tip' class='tip'>"+
							"<div class='title row'>"+
								"<div class='col-xs-6'>"+
									"<p class='tipTitle'>"+Title+"</p>"+
								"</div>"+
								"<div class='col-xs-3'>"+
									"<input type='text' id='tipID' name='tipVal' value='"+HelpfulTipID+"'>"+
								"</div>"+
							"</div>"+
							"<div class='tipDetail row'>"+
								"<p>"+Tip+"</p>"+
							"</div>"+
						"</div>");	
					}else{
					
						tipList.append("<div id='tipMgmt' class='tipMgmt'>"+
							"<div class='row'>");
						if(Active.equals("1")){
							tipList.append("<div class='col-xs-1'><center><i class='fa fa-check active'></i></center></div>");
						}else{
							tipList.append("<div class='col-xs-1'><center><i class='fa fa-times inactive'></i></center></div>");
						}
						
						tipList.append("<div class='col-xs-3'><center>"+DateCreated+"</center></div>"+
								"<div class='col-xs-6'><p class=''>"+Title+"</p></div>"+
								"<div class='col-xs-1'><input type='text' style='display:none' size='1px' id='HelpfulTipID' name='HelpfulTipID' value='"+HelpfulTipID+"'></div>"+
							"</div>"+
						"</div>");
					}				
					;
					
					
					if(DEBUG){
						System.out.println("created record #"+rowCount);
						System.out.println(tipList);
					}
				}
				while(rs != null && rs.next());

然后我在当前页面上包含了新对象。

<div>
	<%=tipList%>
</div>

由于存储的数据是HTML，因此它与ckeditor中输入的数据完全相同！

从SQL HTML数据创建jSON对象

2 个答案: