how to read output of var_export into a variable in PHP?

Question

the output is like below restored in a output.txt file:

array (
  'IMType' => '1',
  'Email' => '[email protected]',
  'SignupName' => 'test11',
  'Password' => '11111',
  'Encrypted' => '',
  'Confirm' => '11111',
  'OldPassword' => '',
  'Name' => 'test',
  'SignupProvinceText' => 'province',
  'SignupCity' => 'cityname',
  'Street' => 'street x.y',
  'SignupIndustry' => 'IT',
  'SignupCompany' => 'jobirn',
  'SignupJt' => 'engineer',
  'CellPhoneNum' => '',
  'linked_in' => '',
)

it's in fact output of var_export(my_variable,true),but how to read it into a variable again?

Samuel L · Accepted Answer · 2023-09-14 16:17:28Z

27

like this:

$dumpStr = var_export($var,true);
eval('$somevar = ' . $dumpStr . ';');

https://www.php.net/manual/en/function.eval.php

edited Sep 14, 2023 at 16:17

Samuel L

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answered Jun 1, 2009 at 3:57

Jonathan Fingland

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8 Comments

Matthew Over a year ago

Be very careful with eval, and only use it if you are 100% completely sure that the user can't inject any code.

Jonathan Fingland Over a year ago

agreed about the danger there. hence my comment to Zilupe's answer.

grawity Over a year ago

Also, it should be eval('$somevar = ' . $dumpStr); as + is only used for addition. (If I remember correctly.)

Jonathan Fingland Over a year ago

hahah, quite right. That's what I get for working on javascript all day

Volodymyr Chumak Over a year ago

Sometimes you should add ; for evaluating variable. For example: eval( '$somevar = '. $dumpStr .';' );

|

jbasko · Accepted Answer · 2009-06-01 03:59:51Z

20

Perhaps you want to serialize object and then unserialize? http://php.net/serialize

answered Jun 1, 2009 at 3:59

jbasko

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4 Comments

omg Over a year ago

the data structure is restored in a file.can it be directly imported without serialization?

Jonathan Fingland Over a year ago

yes, this is probably what he wants. And is safer than using eval

jbasko Over a year ago

@Shore what do you mean by "directly imported without serialization"? Serialization is the process of storing an object. Unserialization is the restoration of the stored object. You can restore objects serialized with anything you wish, but they should follow the format (consult PHP.net)

user Over a year ago

@Jonathan, it is NOT safer than eval. Quote from the doc: Warning Do not pass untrusted user input to unserialize(). Unserialization can result in code being loaded and executed due to object instantiation and autoloading, and a malicious user may be able to exploit this. Use a safe, standard data interchange format such as JSON (via json_decode() and json_encode()) if you need to pass serialized data to the user.

S.Kashizadeh · Accepted Answer · 2022-11-20 20:59:02Z

11

its very simple , you must save it as php file not txt and then

just include it inside a variable :)

like this:

    file_put_contents( '/some/file/data.php', 
                       '<?php return '.var_export( $data_array, true ).";\n" );

when you need to access it just do this :

$data = include '/some/file/data.php'

edited Nov 20, 2022 at 20:59

answered Oct 30, 2018 at 12:57

S.Kashizadeh

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2 Comments

Alexander Garden Over a year ago

includes returns 1 on success and FALSE on failure. Using it that way will not result in the contents of data.php being assigned to $data.

S.Kashizadeh Over a year ago

@AlexanderGarden you are wrong. when you are returning result inside php file you can get value with include inside another variable like what i did for $data. look at 5# example of this link from php.net -> php.net/manual/en/function.include.php

Yay295 · Accepted Answer · 2019-09-09 06:55:30Z

2

This technique is good for data cache:

<?php
// reading data from DB or an API webservice etc.
$arrName = array();    
$arrName = call_procedure_here();

$strFileContent = '<?php' . PHP_EOL . '$arrName = ' . var_export($arrName,true) . ';' . PHP_EOL . '?>';
file_put_contents('cache_folder/arrayfilename.php', $strFileContent);

// later... from another process
include 'cache_folder/arrayfilename.php';
?>

edited Sep 9, 2019 at 6:55

Yay295

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answered Jul 29, 2015 at 5:32

Rody From Argentina

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Comments

Klesun · Accepted Answer · 2019-01-09 15:54:15Z

1

I wrote a basic var_export parser, without eval: VarExportParser.php

Usage:

require_once('VarExportParser.php');
$parsed = VarExportParser::parse("array ('IMType' => '1', 'Email' => '[email protected]')");
print(json_encode($parsed).PHP_EOL);

Outputs:

{"IMType":"1","Email":"[email protected]"}

You can try js version of it here: text_transform.html (the var_export to json button)

answered Jan 9, 2019 at 15:54

Klesun

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