Tell me more ×
Stack Overflow is a question and answer site for professional and enthusiast programmers. It's 100% free, no registration required.
up vote 0 down vote favorite

With PHP, I'm opening a file and it looks like this:

var people = {
vikram: { time1: [ '8:00am', '8:20am', '8:40am', '9:00am',  ], time2: [ '10:20am', '10:40am', '11:00am', '11:20am',  ], time3: [ '8:00am', '8:20am', '8:40am',  ], }};

The variable I'm trying to remove will contain a time (ex. 8:00am) and I will know the timeIndex(ex. time1). I also want to keep all the other times intact.

For example, if I set the variable to 8:40am, I want the new file that is being created to look like this:

var people = {
vikram: { time1: [ '8:00am', '8:20am', '9:00am',  ], time2: [ '10:20am', '10:40am', '11:00am', '11:20am',  ], time3: [ '8:00am', '8:20am', '8:40am',  ], }};

Any help would be appreciated!

share|improve this question
1  
what have you tried...? – Sudhir Apr 30 '12 at 6:46

3 Answers

up vote 0 down vote accepted

you can use preg_replace() for this:

<?php
$filename = 'yourfile.js';
$search = '8:40am';

$content = file_get_contents( $filename );

$pattern = '/(\s*time1:\s*\[.*)([\'"]' . 
           preg_quote($search) .
           '[\'"],?\s*)(.*\])/U';

$content = preg_replace( $pattern, '\1\3', $content ); 

file_put_contents( $filename, $content );
?>

This is a modification of the code example i answered to your last question on a similar topic.

share|improve this answer
what about hello there {$username}! – We are the World Apr 30 '12 at 6:59
for variables inside a double quote, you can enclose them in braces right. – We are the World Apr 30 '12 at 11:12
@JackSpairow and how is that related to this question/answer? – Kaii Apr 30 '12 at 11:32
up vote 3 down vote

The format you show represents a JSON formatted string. You can use json_decode() function to make an array from string, then loop through the array and just unset() the element you don't need.

share|improve this answer
actually that is not valid JSON and can not be easily transformed to valid JSON. – Kaii Apr 30 '12 at 11:35
up vote 0 down vote

Here is the way I did it. Basically, I use json_decode to parse your json to php object. However, I also found that your input is not a well-formed json for php (See example 3). Although my code doesn't look good and generic, but I hope it will help you.

<?php
$json_data = '{
    "vikram": {
        "time1": ["8:00am", "8:20am", "8:40am", "9:00am"], 
        "time2": ["10:20am", "10:40am", "11:00am", "11:20am"], 
        "time3": ["8:00am", "8:20am", "8:40am"]
    }
}';


$obj = json_decode($json_data);

//var_dump($obj->vikram);

$value = "8:40am";
$time1 = "time1";

$delete_item;

foreach($obj->vikram as $name=>$node)
{
    foreach($node as $i => $time)
    {
        if($time==$value && $name=$time1)
        {
            $delete_item = $i;
        }
    }
}

unset($obj->vikram->time1[$delete_item]);
var_dump($obj->vikram);
share|improve this answer
1  
you might check out array_search() – Kaii Apr 30 '12 at 8:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.