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Detect specific info from variable [duplicate]

I have an HTML form field $_POST["url"], having some URL strings as the value.
Example values are:
https://example.com/test/1234?email=xyz#test.com
https://example.com/test/1234?basic=2&email=xyz2#test.com
https://example.com/test/1234?email=xyz3#test.com
https://example.com/test/1234?email=xyz4#test.com&testin=123
https://example.com/test/the-page-here/1234?someurl=key&email=xyz5#test.com
etc.
How can I get only the email parameter from these URLs/values?
Please note that I am not getting these strings from the browser address bar.

You can use the parse_url() and parse_str() for that.
$parts = parse_url($url);
parse_str($parts['query'], $query);
echo $query['email'];
If you want to get the $url dynamically with PHP, take a look at this question:
Get the full URL in PHP

All the parameters after ? can be accessed using $_GET array. So,
echo $_GET['email'];
will extract the emails from urls.

Use the parse_url() and parse_str() methods. parse_url() will parse a URL string into an associative array of its parts. Since you only want a single part of the URL, you can use a shortcut to return a string value with just the part you want. Next, parse_str() will create variables for each of the parameters in the query string. I don't like polluting the current context, so providing a second parameter puts all the variables into an associative array.
$url = "https://mysite.com/test/1234?email=xyz4#test.com&testin=123";
$query_str = parse_url($url, PHP_URL_QUERY);
parse_str($query_str, $query_params);
print_r($query_params);
//Output: Array ( [email] => xyz4#test.com [testin] => 123 )

As mentioned in another answer, the best solution is using parse_url().
You need to use a combination of parse_url() and parse_str().
The parse_url() parses the URL and return its components that you can get the query string using the query key. Then you should use parse_str() that parses the query string and returns
values into a variable.
$url = "https://example.com/test/1234?basic=2&email=xyz2#test.com";
parse_str(parse_url($url)['query'], $params);
echo $params['email']; // xyz2#test.com
Also you can do this work using regex: preg_match()
You can use preg_match() to get a specific value of the query string from a URL.
preg_match("/&?email=([^&]+)/", $url, $matches);
echo $matches[1]; // xyz2#test.com
preg_replace()
Also you can use preg_replace() to do this work in one line!
$email = preg_replace("/^https?:\/\/.*\?.*email=([^&]+).*$/", "$1", $url);
// xyz2#test.com

Use $_GET['email'] for parameters in URL.
Use $_POST['email'] for posted data to script.
Or use _$REQUEST for both.
Also, as mentioned, you can use parse_url() function that returns all parts of URL. Use a part called 'query' - there you can find your email parameter. More info: http://php.net/manual/en/function.parse-url.php

You can use the below code to get the email address after ? in the URL:
<?php
if (isset($_GET['email'])) {
echo $_GET['email'];
}

I a created function from Ruel's answer.
You can use this:
function get_valueFromStringUrl($url , $parameter_name)
{
$parts = parse_url($url);
if(isset($parts['query']))
{
parse_str($parts['query'], $query);
if(isset($query[$parameter_name]))
{
return $query[$parameter_name];
}
else
{
return null;
}
}
else
{
return null;
}
}
Example:
$url = "https://example.com/test/the-page-here/1234?someurl=key&email=xyz5#test.com";
echo get_valueFromStringUrl($url , "email");
Thanks to #Ruel.

$web_url = 'http://www.writephponline.com?name=shubham&email=singh#gmail.com';
$query = parse_url($web_url, PHP_URL_QUERY);
parse_str($query, $queryArray);
echo "Name: " . $queryArray['name']; // Result: shubham
echo "EMail: " . $queryArray['email']; // Result:singh#gmail.com

A much more secure answer that I'm surprised is not mentioned here yet:
filter_input
So in the case of the question you can use this to get an email value from the URL get parameters:
$email = filter_input( INPUT_GET, 'email', FILTER_SANITIZE_EMAIL );
For other types of variables, you would want to choose a different/appropriate filter such as FILTER_SANITIZE_STRING.
I suppose this answer does more than exactly what the question asks for - getting the raw data from the URL parameter. But this is a one-line shortcut that is the same result as this:
$email = $_GET['email'];
$email = filter_var( $email, FILTER_SANITIZE_EMAIL );
Might as well get into the habit of grabbing variables this way.

$uri = $_SERVER["REQUEST_URI"];
$uriArray = explode('/', $uri);
$page_url = $uriArray[1];
$page_url2 = $uriArray[2];
echo $page_url; <- See the value
This is working great for me using PHP.

In Laravel, I'm using:
private function getValueFromString(string $string, string $key)
{
parse_str(parse_url($string, PHP_URL_QUERY), $result);
return isset($result[$key]) ? $result[$key] : null;
}

A dynamic function which parses string URL and gets the value of the query parameter passed in the URL:
function getParamFromUrl($url, $paramName){
parse_str(parse_url($url, PHP_URL_QUERY), $op); // Fetch query parameters from a string and convert to an associative array
return array_key_exists($paramName, $op) ? $op[$paramName] : "Not Found"; // Check if the key exists in this array
}
Call the function to get a result:
echo getParamFromUrl('https://google.co.in?name=james&surname=bond', 'surname'); // "bond" will be output here

echo statement not working? Seems to be formatted correctly

I'm creating a object in javascript echo'd by PHP, the syntax seems to be correct but it is giving me an error.
The variables I am passing to the object are strings and I have tried putting them in quotes, etc. I also tried creating a JS function to do this and it had similar results.
<?PHP
$TEDS = $db->query("
SELECT a.*
FROM TEDS a
");
//Add each TED location into a javascript array.
echo '<script type="text/javascript">';
while($TED = $TEDS->fetch_assoc())
{
echo 'var pos = {lat: '.$TED['latitude'].', lng: '.$TED['longitude'].'};'; //THIS ECHO STATEMENT IS THE ONE GIVING ME PROBLEMS.
}
echo '</script>';
?>

It's highly likely that one of your latitude values does not have a value, causing it to print something like the following:
var pos = {lat: , lng: 1234}
Without a value to go with the property, the JavaScript parser would unexpectedly see the comma. This would greatly explain an unexpected comma syntax error.
So, first of all, you do NOT manually construct JSON. PHP has json_encode() that converts PHP arrays into JSON. Use that to correctly build JSON for you.
Next, your code prints pos repeatedly, overriding the previous values. If you only care about the last value, don't print everything. If you do care about all the values, store them in an array:
<?php
// Get your TEDS
$teds = $TEDS->fetch_assoc();
// Using array_map to convert the items from latitude-longitude to lat-lng.
$teds_array = array_map(function($ted) {
return ['lat' => $ted['latitude'], 'lng' => $ted['longitude']];
}, $teds);
// Serialize to JSON
$teds_json = json_encode($teds_array);
?>
// positions will be an array of objects with lat-lng.
var positions = <?php echo $teds_json; ?>
Lastly, be careful of XSS. Do not just print unsanitized values like above. Make sure your data is clean before printing it on the page.

i would try to write the full text, after the loop. This way you can build a full json array:
<?PHP
$TEDS = $db->query("
SELECT a.*
FROM TEDS a
");
//Add each TED location into a javascript array.
echo '<script type="text/javascript">';
$text = 'var positions = [';
while($TED = $TEDS->fetch_assoc()) {
$text .= '{lat: '.$TED['latitude'].', lng: '.$TED['longitude'].'},';
}
// remove last comma
$text = substr($text,0 , strlen($text)-2);
$text .= '];'
echo $text;
echo '</script>';
?>
Another way would be to just json_decode($TED).

Parse Laravel JSON object in Javascript throwing error

Laravel code :
$teachers = Teachers::where('possessed_by_community', $communityId)->pluck('teacher_name');
return view('pages.show_add_teachers', [
'teachers' => $teachers
]);
Then in client side I tried :
var teachers = "<?php echo json_encode($teachers) ?>" ;
teachers = JSON.parse(teachers);
console.log(teachers);
In webconsole I get :
SyntaxError: missing ; before statement
In firefox debugger :

Use single quotes instead var teachers = '<?= json_encode($teachers) ?>';
or don't use quotes at all, leaving out JSON.parse(teachers);.
var teachers = <?= json_encode($teachers) ?>;
console.log(teachers);
Though you must check $teachers is valid, string, null or array at all times.
Else it would become var teachers = ; and break.

var jArray = #json($teachers);

PHP base64_decode() .toDataURL()

I have an encoded string created from the Javascript function .toDataUrl(). Every time I try to convert this to binary with PHP base64_decode() it truncates the binary. I have already tried various items such as the following:
//First Attempt
$encoded = str_replace('data:image/png;base64,', '', $sig);
$decodedstring = base64_decode(str_replace(array(' ', '_'), array('+', '/'), $encoded));
$decodedstring = base64_decode(chunk_split($encoded));
//2nd attempt
$encoded = str_replace([' ','data:image/png;base64,'], ['+',''], $sig);
$decodedstring = "";
for ($i=0; $i < ceil(strlen($encoded)/256); $i++)
$decodedstring = $decodedstring . base64_decode(substr($encoded,$i*256,256));
//Other attempt
$decodedstring = base64_decode( str_replace(['data:image/png;base64,', ' '], ['','+'], $sig) );
None of these produce the correct file. Also to note, that when I use the standard png base64 online decoders the image looks 100% correct so the issue seems to be happening in the conversion.
Would anyone have additional thoughts. I have spent a few days trying to research this with no luck.

This is a function I made to convert data sent via ajax to php. It has always worked flawlessly for me.
function prepare_image($string, $file_name) {
list($type, $string) = explode(';', $string);
list(, $string) = explode(',', $string);
$string = base64_decode($string);
return $string;
}

json data value type

I want to change json array values from strings to float values which from php script. It's all string type. I want it as a float type.
[{"year":"2008","value":"4169.20"},
{"year":"2009","value":"4067.50"},
{"year":"2010","value":"4848.40"},
{"year":"2011","value":"5654.80"},
{"year":"2012","value":"6071.50"}]
But i want it to look like. Is it possible?
[{"year":"2008","value":4169.20},
{"year":"2009","value":4067.50},
{"year":"2010","value":4848.40},
{"year":"2011","value":5654.80},
{"year":"2012","value":6071.50}]

The simplest way is to cast value to float before encoding data to the JSON.
But if you want to work only in JS:
var data = [ /* your data */ ];
data.forEach(function(item){
item.value = +item.value;
});
Output:
[
{"year":"2008","value":4169.2},
{"year":"2009","value":4067.5},
{"year":"2010","value":4848.4},
{"year":"2011","value":5654.8},
{"year":"2012","value":6071.5}
]

<?php
$data = '[{"year":"2008","value":"4169.20"},
{"year":"2009","value":"4067.50"},
{"year":"2010","value":"4848.40"},
{"year":"2011","value":"5654.80"},
{"year":"2012","value":"6071.50"}]';
$a = json_decode($data);
$b = array();
foreach ($a as $key => $v) {
$c['year'] = $v->year;
$c['value'] = (float) $v->value;
$b[] = $c;
}
echo '<pre>';
print_r(json_encode($b));
echo '</pre>';
?>
Demo

use javascript parseFloat() function:
e.g.
var b = parseFloat("10.00")

I see only JS answers. I don't know what is your intention, but if you have the jsong string in PHP you can use something like this:
$json = ' [
{"year":"2008","value":"4169.20"},
{"year":"2009","value":"4067.50"},
{"year":"2010","value":"4848.40"},
{"year":"2011","value":"5654.80"},
{"year":"2012","value":"6071.50"}
]';
$jsonArray = json_decode($json);
foreach($jsonArray as $key => &$subArray)
{
$subArray->value = (float) $subArray->value;
}
echo json_encode($jsonArray);

You have 2 options.
Option 1.
In the backend of your application (the php code) you can cast "value" to be float. E.g.:
$floatValue = (float) $strValue;
Option 2.
If you can't edit the backend use parseFloat on each "value". It will produce floats from the strings in your response.

with preg_replace
$json = preg_replace('/"value":"(\d+\.\d+)"/', '"value":$1',$json);