Could anyone help me pleaseeeee
Dear OlegK,
Thak you for your response. I found this guide from tony:
$(“#mygrid”).jqGrid({
…
url: “myurl.php”
datatype: “local”,
…
});
$(“#mybutton”).click(function(){
Dear Phil,
Yoou are awesome. It works. It helped me so much.
Thank you very much.
Best Regards,
Candra
Dear all,
You gave an example that “inv no” column content is the basename of an image called “.jpg” file.
So, the content “1″ will show in a tooltip the “1.jpg image”, “2″ shows in a tooltip the “2.jpg” image and so on…
How if I want to have a thumbnail based on primary ID of a row?
for example a row with primary ID = xxx will show in a tooltip the “xxx.jpg image” and so on…
Best Regards,
Candra
Dear ppr,
Thank you for your guide to jquery qtip. I was success to follow your step but I got problem when I want to change thumbnail tooltip when the cursor is on “Tax” column (fifth column).
I have changed the code to:
var col1= $(this).find(“td:eq(4)”); // Fifth column Tax
and the image did not show.
Could you help me solve this problem, please? Thank you
Best Regards,
Candra
Dear all,
I have tried to use tooltip above and did not success, does anybody can guide me with a working sample tumbnail image hover pease?I have a project with a lot of goods and i need tumbnail images to make it easy.
Best regards,
Candra
Dear all,
I have tried to use tooltip above and did not success, does anybody can guide me with a working sample tumbnail image hover pease?I have a project with a lot of goods and i need tumbnail images to make it easy.
Best regards,
Candra
Try this one
$.ajax({
I have found the way.
The problem has solved.
Dear all,
I have changed my validate function to this:
function usercheck(value,ID){
Dear all,
Could anybody assist me how to solve server validation on dynamic editing, please?
I want server to validate user if someone create double ID.
I have been trying for searching this problem from this forum but i did not find the clue.
Please… I really need your help
Regards,
Candra
Dear ddotsenko,
Thank you for your clue. my grid has solved.
Regards,
candrairawan78
Hi there,
I made linkable row to show a chart with showlink formatter.
It shows Chart.phpshow?id=null, I want to change to Chart.php?id=B86300-BZ010, where B86300-BZ010 is row id value. What should I do to erase the show and fill the id ?
If i used link formatter, it will direct link to B86300-BZ010.
Could you help me please,
Candra
Dear Tony,
Thank you for your responce.
OK, I understand and appreciate your forum.
I have changed reloadAfterSubmit to false , but it only delete rows in grid, not delete data in MySQL.
Could you suggest where I can get help for my problem?
Regards,
Candra
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