如何通过javascript在表字段上显示值? [英] How to display a value on the table field through javascript?

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问题描述

Down是我的代码,用于从项目表中选择活动项目的表中列出特定客户的名称。



 function addTable(){

var myTableDiv = document.getElementById( myDynamicTable);
var table = document.createElement(' );
table.border = ' 1';

var tableBody = document.createElement(' TBODY');
table.appendChild(tableBody);
db.transaction(函数(事务)
{

var pro = localStorage.getItem( ActiveProject);
var activePro = ' 从ProjectTable中选择Prjdb_id,其中Proj_name =' + pro + ' ';
transaction.executeSql(activePro,[],function(transaction,result)
{
for var i = 0 ; i < result.rows.length; i ++)
{
var row = result.rows.item(i);
id = row.Prjdb_id;
}
db = openDatabase(id,version,displayName,maxS IZE);
if (id)
{
transaction.executeSql(' SELECT * FROM customer',[],function(transaction,result)
{

le = result.rows.length;
if (result.rows.length!= 0
{
var row = result.rows.item(i);
var name = row .Cust_name;
for var i = 0 ; i< 2; i ++)
{
var tr = document.createElement(' TR');
tableBody.appendChild(tr);
F或 var j = 0 ; J<文件; j ++)
{
var td = document.createElement(' TD');
td.width = ' 75';
td.appendChild(document.createTextNode(name));
tr.appendChild(td);
}
}
myTableDiv.appendChild(table);
}
},errorHandler);
}
});
});







 function errorHandler(tx ,错误)
{
alert( 数据库错误: +错误) ;
}





一直运行到if(id)状态,直到它打开的状态数据库也没有。但是没有进入执行sql命令的下一行。

任何人都可以帮助我......

解决方案

< pre lang =Javascript> db1 = openDatabase(id,version,displayName,maxSize);

db1.transaction( function (transaction)
{
/ / 你的代码在这里
});





代码内部按上述方式进行更改。

由于这里使用了两个数据库,第一个是主数据库,第二个是项目数据库第二个数据库应该有不同的名称。



这对我来说很有用..







更正格式问题。

添加标签。

[/编辑]


Down is my code to list the names of a particular customer from a table selecting the active project from the project table.

function addTable() {

    var myTableDiv = document.getElementById("myDynamicTable");
    var table = document.createElement('TABLE');
    table.border='1';

    var tableBody = document.createElement('TBODY');
    table.appendChild(tableBody);
  db.transaction(function(transaction)
    {
    
     var pro=localStorage.getItem("ActiveProject");
     var activePro='Select Prjdb_id from ProjectTable where Proj_name="'+pro+'" ';
    transaction.executeSql(activePro , [],  function(transaction, result)
    {
             for (var i = 0; i <  result.rows.length; i++)
             {
                var row = result.rows.item(i);
                    id=row.Prjdb_id;
             }
               db=openDatabase(id, version, displayName,maxSize);
      if(id)
         {
          transaction.executeSql('SELECT * FROM customer' , [],  function(transaction, result)
          {
         
            le=result.rows.length;
            if (result.rows.length != 0)
            {
            var row = result.rows.item(i);
            var name= row.Cust_name;
            for(var i=0;i<2;i++)
            {
            var tr = document.createElement('TR');
            tableBody.appendChild(tr);
            for (var j=0; j<le; j++)
            {
            var td = document.createElement('TD');
            td.width='75';
            td.appendChild(document.createTextNode(name));
            tr.appendChild(td);
            }
            }
            myTableDiv.appendChild(table);
            }
          },errorHandler);
         }
    });
 });




function errorHandler(tx, error)
{
  alert("Database Error: " + error);
}



It goes well till the if(id) condition inside the condition it opens the database too.But doesnot enter the next line of executing the sql command.
Can anybody please help me...

解决方案

db1 = openDatabase(id, version, displayName,maxSize);

db1.transaction(function(transaction)
{ 
    //ur code here 
});



Inside the code make changes as the above.
Since there are two databases used here the first one is Main database and the second is Project database second db should has a different name.

This worked well for me..


[Edit member="Tadit"]
Corrected formatting issues.
Added pre tags.
[/Edit]


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