Update DB field using Zend Framework

Question

I'm a n00bie with Zend but I'm learning it. I want to update a field in the database using a button, but I don't know how to do it.

This is what I wanna do: UPDATE $table SET content_field=1 WHERE $id = contentId

For example if I press that button I wanna put a 1 in the field instead of a default 0.

max4ever · Accepted Answer · 2012-01-26 10:23:46Z

1

Assuming you have a Client table and created the model

$model_client = new Model_Client();
$client = $model_client->find($my_client_row_id)->current();
$client->some_field = 1;
$client->save();

answered Jan 26, 2012 at 10:23

max4ever

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ianaz · Accepted Answer · 2012-01-26 10:42:45Z

0

You can do an update query with this code

$db = Zend_Db_Table::getDefaultAdapter();
$table = new Application_Model_DbTable_TableName();
$where = $table->getAdapter()->quoteInto('id = ?', $contentId);
$data = array('content_field' => 1);
$table->update($data, $where);

edited Jan 26, 2012 at 10:42

answered Jan 26, 2012 at 10:30

ianaz

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5 Comments

Mark Over a year ago

sorry for the stupid question, but can I link to the button? Just because with PHP, I usually do something like that: <form name="form1" method="post" action="testForm.php"> <input type="Submit" name="Submit1" value="Update"> </form> and in the PHP file I write the sql actions, but I still don't understand how can I link the button from my view to a real action.

ianaz Over a year ago

If you want to enter in the method "updateDataAction" in the controller "User" you need to do the post to /user/updatedata and access the post data with $this->_request->getParam('INPUT_NAME_PARAM'); Is that what you meant?

Mark Over a year ago

I hard coded a form in view/content/myfile.tpl like this: <form method="post" action="/Admin/"/> <select name="content_field"> <option value=""></option> <option value="0">Yes</option> <option value="1">No</option> <input value="Go" type="submit" /> </form> But I don't know how to "link" that submit button to an Update DB field function like this one: UPDATE $table SET content_field=1 WHERE $id = contentId

ianaz Over a year ago

You have to create a controller ( framework.zend.com/manual/en/zend.controller.quickstart.html ) with inside an action (framework.zend.com/manual/en/zend.controller.action.html ). Your form should then link to /namecontroller/nameaction and in the method of the action you have to get the posted value (content_field) and execute the code i posted above

Mark Over a year ago

Oh cool, thank you very much. I'll read up well the pages and hope to fix the problem. Thank you very much for the help.

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