Regex not working properly in PHP

Question

I have this data:

{"names":["George","Eric","Alice"]}

And I want to use preg_match_all to filter out the words in between quotation like this:

$s = $data;

if (preg_match_all('/"([^:]*)"/', $s, $matches)) {
        echo join("\n", $matches[1]);
}

But this is outputting names George","Eric","Alice I tried many things but I cant figure it out.

json_decode() to the rescue print_r(json_decode('{"names":["George","Eric","Alice"]}', true)); ! — HamZa
– HamZa, Commented Jan 30, 2014 at 14:08
@Youss It's very clear that this is just to show what the array will contain. You only have to call the elements of this array or just use join(). — HamZa
– HamZa, Commented Jan 30, 2014 at 14:14
@Youss ok, you haven't mentioned that beforehand. But you could use a foreach($arr, $key => $value){} loop, it actually depends on how random your source is. Anyways, parsing JSON with regex isn't "recommended" it's still possible with regex though — HamZa
– HamZa, Commented Jan 30, 2014 at 14:25

falsetru · Accepted Answer · 2014-01-30 14:18:55Z

3

* matches greedy (as much as possible). Use non-greey version: *?

if (preg_match_all('/"([^:]*?)"/', $s, $matches)) {
    echo join("\n", $matches[1]);
}

output:

names
George
Eric
Alice

UPDATE

json_decode is more appropriate for this kind of work. Try following:

foreach (json_decode($s, true) as $key => $value) {
    echo $key . "\n";
    echo join("\n", $value);
}

edited Jan 30, 2014 at 14:18

answered Jan 30, 2014 at 14:13

falsetru

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7 Comments

user2169391 Over a year ago

Don't treat json data like strings! Bad practice.

falsetru Over a year ago

@Roebie, I updated the answer. I just wanted to tell the reason why OP get the result.

user2169391 Over a year ago

@falsetru: Sorry, our comments crossed.

Youss Over a year ago

You have the variable 'names' This is the problem, Im scraping the JSON and I dont know the variable upfront, the JSON is not mine and the data keeps changing

HamZa Over a year ago

Actually you need a complexer regex to take into account escaped double quotes (?<!\\)"((?:[^\\]|\\.)*?)"

|

anubhava · Accepted Answer · 2014-01-30 14:08:53Z

2

This is actually JSON string use json_decode to parse it rather than using regex on this:

print_r(json_decode('{"names":["George","Eric","Alice"]}', true));

OUTPUT:

Array
(
    [names] => Array
        (
            [0] => George
            [1] => Eric
            [2] => Alice
        )

)

answered Jan 30, 2014 at 14:08

anubhava

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2 Comments

Youss Over a year ago

The thing is Im outputting this as plain text not as json. I really just need the words not the quotation, brackets and so on. (my project requires it to be plain text..)

anubhava Over a year ago

Then json_decode can be used as: print_r(json_decode($data, true)); where $data is your variable containing this plain text. That is much cleaner and safer than regex.

Arrok · Accepted Answer · 2014-01-30 14:16:03Z

1

try this

$strContent = '{"names":["George","Eric","Alice"]}';
$strRegex = '%\"(.+?)\"%s';
if (preg_match_all($strRegex, $strContent, $arrMatches))
{
    var_dump($arrMatches[1]);
}

answered Jan 30, 2014 at 14:16

Arrok

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Comments

score 1 · Accepted Answer · 2014-01-30 14:27:10Z

1

Since your data is json formatted you should really treat is as json and not process it with regex which is to be used for strings. Try this:

$json = '{"names":["George","Eric","Alice"]}';
$data = json_decode($json, true);
foreach($data['names'] as $item) echo "$item\n";

Or without the hard coded "names":

$json = '{"names":["George","Eric","Alice"]}';
$data = json_decode($json, true);
foreach($data as $arr) foreach($arr as $item) echo "$item\n";

edited Jan 30, 2014 at 14:27

answered Jan 30, 2014 at 14:16

user2169391

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