Insert a element from php variable in associative array

Question

I have a conditional string like this:

echo $cnd = "'use_funds'=>'no'";

And my $data array is :

$data = array(
    $cnd,
    'quantity'   => $_POST['qty_' . $q],
    'veg_name'   => $_POST['veg_name_' . $q],
    'rate'       => $_POST['rate_' . $q],
    'ws_price'   => $_POST['ws_price_' . $q],
    'ret_price'  => $_POST['ret_price_' . $q],
    'mop'        => $_POST['mop_' . $q],
    'ret_margin' => $_POST['ret_margin_' . $q]
);

The above echo $cnd shows 'use_funds'=>'no', but the var_dump($data) shows array 0 => string ''use_funds'=>'no'' (length=17) . And since its a string my data is not inserting in my table. How can I make the $cnd to a associative array element instead of string?

Why does $cnd contain PHP code? Wouldn't it make more sense to have something like $use_funds = 'no'? — Bailey Parker
– Bailey Parker, Commented Jun 18, 2013 at 6:31

RST · Accepted Answer · 2013-06-18 06:23:35Z

3

if (your condition) {
  $data['use_funds'] = 'no';
} else {
  //some other code
}

answered Jun 18, 2013 at 6:23

RST

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1 Comment

Akilan Over a year ago

Hi Nithish, if you got solution, then accept RST answer. Dont open this question

Code Lღver · Accepted Answer · 2013-06-18 06:23:41Z

2

don't use the $cnd part to make and insert the data in the array:

Use the below code:

$data['use_funds'] = 'no';

And this code will append the array. OR you can use the array_push function of PHP.

answered Jun 18, 2013 at 6:23

Code Lღver

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3 Comments

Bailey Parker Over a year ago

Note that array_pushing an associative array doesn't work as you'd expect (codepad.org/b4of4dml)

Code Lღver Over a year ago

@PhpMyCoder ok, Then I can suggest array_merge by making two array and merge them.

Bailey Parker Over a year ago

Sounds better. +1 on the suggestion.

DivinusVox · Accepted Answer · 2013-06-18 06:41:42Z

1

If it must be a string, split it and add it to the array. http://php.net/manual/en/function.explode.php

The explode function takes a string and breaks it into an array

$my_string_split = explode('=>', $cnd);
// returns [0 => 'use_funds', 1 => 'no']
$data[$my_string_split[0]] = $my_string_split[1];

If it does not need to be a string, do it literally:

$data['use_funds'] = 'no';

edited Jun 18, 2013 at 6:41

answered Jun 18, 2013 at 6:35

DivinusVox

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Comments

Ripa Saha · Accepted Answer · 2013-06-18 06:28:30Z

first of all your $cnd is string , not an array. and you are trying to append this $cnd with $data array.

for adding element to an array we basically use push() method. For your case if you use array_push method then you will get a output like below

array_push($cnd,$data);
$data=array(
                                    0=>'use_funds'=>'no',
                                    'quantity'=>$_POST['qty_'.$q],
                                    'veg_name'=>$_POST['veg_name_'.$q],
                                    'rate'=>$_POST['rate_'.$q],
                                    'ws_price' => $_POST['ws_price_'.$q],
                                    'ret_price' => $_POST['ret_price_'.$q],
                                    'mop' => $_POST['mop_'.$q],
                                    'ret_margin' =>$_POST['ret_margin_'.$q]
  );

so my suggession will be if you want you desire result then follow below step:-

 either define your $cnd as an array and then use array_merge();

Kevin · Accepted Answer · 2013-06-18 06:30:22Z

0

echo $cnd = "'use_funds'=>'no'";

The above line considers "'use_funds'=>'no'" as a string and assigns it to $cnd. So no matter what, it always remains as a string unless you make any proper change to it. The default key in that case will be 0. So, it will be like:

[0] => "'use_funds'=>'no'"

To get around this problem, you can do this:

$cnd = array('use_funds'=>'no');

Then, you an use array_merge() function to merge the two like this:

array_merge($cnd, $data);

answered Jun 18, 2013 at 6:30

Kevin

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Comments

DevZer0 · Accepted Answer · 2013-06-18 06:24:47Z

-1

You can accomplish it using eval

$cnd = "'use_funds'=>'no'";

eval("\$x = array($cnd);");

print_r($x);

answered Jun 18, 2013 at 6:24

DevZer0

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2 Comments

DivinusVox Over a year ago

Please don't use eval, it attempts to interpret the given string as code and as such is very unsafe, specifically in conditions where the string is not known to be good. Best to just leave it be, regardless, there's almost always a safer way to get the result.

DevZer0 Over a year ago

eval has it's uses. this situation is a pre-defined input so there is no safety concern. my solution didn't change any of his current setup. Everyone else assumes that he didn't know about the usual array key addition.

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