how can display items of a hidden colmodel when mouse over another column which shows only some part of the data. in user column i have many names, i only display first user name and rest the count. so i passed the full names as hidden. I want to display that names according to the mouseover
If I correctly understand the problem you can solve it by usage cellattr callback which defines title attribute based on the value of another (hidden) column. See the answer for the corresponding code example.
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Forgive my ignorance for the proper terminology, but let me try to explain what I want to happen.
I have two custom fields on the incident table/form. I have created a custom table with 2 columns. I have figured out how to reference the table in one of the fields, allowing me to search the entries. Now I want to link the field selection to the other field via the custom table I made. When I make a selection in field A, I want field B to populate with the other column on the same row Field A pulled from in the first place. How would I do that?
Sounds like you want what's commonly referred to as a "derived field" or a dot-walked field.
You have a reference field which stores the reference to the other table, and want a second field on the form that shows another field on that referenced table. You don't actually need to create a new element, you just need to add a dot-walked form element.
Once you have the reference field added, go to Personalize/Configure the form layout.
In the slushbucket of available fields, you'll notice that reference
fields show up in green text with a little [+] next to them.
Select your reference field and a little button will show up between the two
lists, just above the "Add" button
Click that button and the left-side available fields will show the fields available on that reference field's table.
From here, select that second field that you want to display on your form, and bring it over to the right side where you want it.
I would like to pass a search value to one of the columns in the jsgrid. Can this be done?
Say I have a grid with the following columns:
City, Size, and Postal Code
I would like to pass in a postal code value to display in the toolbar search box and then run the search.
You can access the text fields of the searching toolbar by ids. The id of the filed will be build based on the value of name property of the column in colModel and the prefix "gs_". For example if you have columns with name="city" then to set the value "London" in the corresponding input field you can use $("#gs_city").val("London"). After you set all the values which you need you can trigger change event on any of the elements of the toolbar to apply the filtering on the grid. Alternatively one can call triggerToolbar method saved on DOM element of the grid: $("#gridid")[0].triggerToolbar();.
JQGrid Form edit input screen, Can u give some idea on how to hide the selected area and also to display it on click of the check box.
for example
my form editing window have
show full details check box and First, Last Name, Age, and Address, Zipcode, City, State and Country
if that check box is checked First, Last Name, Age, and Address, Zipcode, City, State and Country fields are shown else all fields are hidden
Is it possible ?
If I understand correct your question you want test the value of some field from the Edit form and hide some other fields of the form depend on the tested value.
You can implement the behavior inside of beforeShowForm callback. If you you have for example column with name: "sold" in the colModel which has formatter: "checkbox" then you will find the corresponding value in the edit form in the checkbox having id="sold". So you can use $("#sold").is(":checked") to test the value. You can do such testing inside of beforeShowForm callback which will be called after the form will be initialized, but before it will be shown. To hide information from "address" column for example you can use $('#tr_address').hide();. The $('#address') represent the input field for the address and $('#tr_address') represents the full row of the edit form with the information.
You can find and example which very close to what you need in the answer.
I need to show/hide columns in jqGrid and to filter grid results.
For hiding and showing I use global array on the page. It keeps state.
For searching and filtering I use build-in filter dialog.
For invisible columns I set property "search" for "false"(nobody wants to sort results if he doesn't see it) so...there is no such column in the filter list. But when I show any column that was hidden(I call $("#jqgrid").showCol(colName);) then this column doesn't appear in the filter list. Of course, I have to change column's "search" property. But how??? Is it possible?
Thanks in advance
I've got an autocompleter in a CakePHP app that provides a dropdown list of neighborhoods to attach to a place. I'm pulling from a table 'neighborhoods' with name and id fields. I want the user to see the name of the neighborhood, and I want the app to see the id of the neighborhood and insert it into the places table. How can I make this happen?
Another way of doing this would be to storing the id. Then when a user presses submit you replace the input field's value with your stored value.
assign the correct (hidden) value to a hidden input field
This type of functionality is normally achieved with a combobox, which is like an autocomplete, with an underlying select elememt, see here