CakePHP MySQL Values

Question

echo $this->Form->input('quantity', array('options' => array('1','2','3','4')));

I have this code to input the values 1,2,3 or 4 into the quantity field in my database.

If a user selects 1 it inputs 0, 2 inputs 1 etc.

If the code is

echo $this->Form->input('quantity');

The user can input what they like and guess what... It works. What am I doing wrong?

thecodeparadox · Accepted Answer · 2013-03-05 18:43:05Z

1

I think you're trying to implement a drop-down box. I think you can try this:

$options = array('1' => '1', '2' => '2', '3' => '3', '4' => '4');
echo $this->Form->input('quantity', array('options' => $options, 'default' => '1'));

this will create a select drop down box.

For detail please see here.

answered Mar 5, 2013 at 18:43

thecodeparadox

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4 Comments

blairmeister Over a year ago

Thanks. This worked. I had drop down box, but it thought each number was 1 less. Does the '1'=>'1' etc fix it?

thecodeparadox Over a year ago

yes. if you put only options array then it will index those array item like normal array mechanism. But for your custom indexing you need to set options as associative array (key => value) pair.

thaJeztah Over a year ago

The difference with the original code, is that CakePHP uses the 'keys' (indexes) of the array as values and the 'values' as visible labels; array(1,2,3); is equal to array(0 => 1, 1 => 2, 2 => 3); i.e. the value of 1 is 0

thecodeparadox Over a year ago

@BlairMurray welcome.. actually we all are learning every moment.

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