Here is my SQL statement:

SELECT  Cast(t1.ProposalID as varchar) as proposalid,  t1.proposalno, t1.proposalsur

….

The proposalid is an int field, but as far as I can tell … it should be returned as a varchar.  

The data does display correctly in the grid, but not as a link.

If i change the SQL statement to be:

SELECT 'a' + Cast(t1.ProposalID as varchar) as proposalid,  t1.proposalno, t1.proposalsur

….

then the link appears and works as expected.  (However the link does not work because of the added 'a')

I have, what I believe is the latest version downloaded …. jquery.jqGrid-3.6.5

Any help would be appreciated.

Fixed. Best Regards

Tony

tony said:

Hello,

The code above tell you why!

if(isString(cellval))

....

Will fix it.

Best Regards

Tony

Tony, has this fix been rolled into any downloadable file?

The code above tell you why!

if(isString(cellval))

....

Will fix it.

Best Regards

Tony

It works fine as ling as the cell value is string. If the cell value is number, it does not show link.

I used the console.log( $("#mygrid").getGridParam('colModel') ); option and it is displaying all parameters correctly.

Any idea?

Thanks

@Johannes

The only thing if I see the code is - if your value a string. We make this checking for show link

@newperson

You can at write your own custom formatter if the provided does not meet your needs.

Best Regards

Tony

Everything was going good until I wanted to use data inside my cells.

My question is: I want to use data inside my cell to put on the link, is the showlink restricted to id fields only, it seems to be hard coded and there isn't an option to specify anything regarding posting cell values. Would I need a custom formatter to do this?

I was able to get the basic example to work but it's using the id not other cell values

-> http://ubuntu/openaudit/system.....ew=summary

Now I have service tags that I'd like to link to the dell website, however these service tags aren't id's they are cell values.

I was not able to get this to work it returns

->http://support.dell.com/suppor.....4257200001

*SW4257200001 is data retrieved using json and this is the id field which I don't need

system_id_number is:

Thanks and hope this makes sense.

Best Regards

Johannes Geppert

Everthing seem to be OK, so I think that something is not loaded. The formatter does not depend on any other module

When you have the grid please check at the top which modules do you have.

Best Regards

Tony

Could you please dump the colModel after the grid is created, and see what you have for this column

comething like (using FireBug)

Regards

Tony

also when I use the link formatter.

But when I use the showlink formatter I got only a cell with value but it is not

rendered as Link.

I try it with various combinations of formatoptions but no way.

Are there any dependencies to use this formatter?

Thanks for your help

Johannes

I'm not sure what you do, but there is a demo with such example and it works ok.

What does not work actually?

Regards

Tony

formatoptions = { baseLinkUrl : '/jquery-grid/grid.action', target : '_new' };

But without Success. Are there any prerequistes for showlink formatter?

I have read your Example, why there is set the formatter to "select"instead of "showlink" in the Wiki?

replave the values of baseLinkUrl and showAction

Good reading of docs sometimes help 

http://www.trirand.com/jqgridw.....nk_example

I try to format a Column as Link. I try it with different Options but it never renders a Link for me.

employeenumber = {};

employeenumber.name = "salesemployee.employeenumber";

employeenumber.jsonmap = "salesemployee.employeenumber";

employeenumber.index = "salesemployee.employeenumber";

employeenumber.formatter = "showlink";

employeenumber.formatoptions = { baseLinkUrl : '', showAction : '/jquery-grid/grid.action', addParam : '&action=show', target : '_new' };

employeenumber.editable = false;

gridloadtable_colnames.push("Employee");

gridloadtable_colmodels.push(employeenumber);

Does anyone have an suggestion?

Johannes