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?
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(' 表跨度>);
table.border = ' 1';
var tableBody = document.createElement(' TBODY'跨度>);
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'跨度>);
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......
< 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。
添加预
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...
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 isMain database
and the second isProject database
second db should has a different name.
This worked well for me..
[Edit member="Tadit"]
Corrected formatting issues.
Addedpre
tags.
[/Edit]
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