How to properly add a "+1" to the php variable [closed] - php

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I am a little lost with one of my variables and need your help. I am trying to add the +1 to default to my phone number variable so I don't have to use it on front.
Current row:
$this->sendSMS($plainMessage, $users['mobile_number']);
What I tried is next:
$mobile_number = "+1" . $mobile_number
$this->sendSMS($plainMessage, $users['mobile_number']);
I am pretty sure it's wrong but my brain is stuck.

Option 1:
$mobile_number = "+1" . $users['mobile_number'];
$this->sendSMS($plainMessage, $mobile_number);
Option 2:
$this->sendSMS($plainMessage, "+1" . $users['mobile_number']);
Spaces before and after dot are not necessary.

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php preg_replace working only if there's ending space in string [closed]

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Closed 7 years ago.
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I have this pattern:
preg_replace_callback('##abc\((.*?)\)(.*?)#end.#is', ..
My template string:
$test = "#abc('test')<h1>test</h1>#end"; // not working
$test2 = "#abc('test')<h1>test</h1>#end "; // working
Why it doesn't work if there's no space after #end?
As #Rizier123 pointed out, this is the correct regex:
preg_replace_callback('##abc\((.*?)\)(.*?)#end#is', ..

What is the "returnto" in this piece of code [closed]

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header("Location: {$TBDEV['baseurl']}/login.php?returnto=" . urlencode($_SERVER["REQUEST_URI"]));
Is it a variable? PHP reserved word? something to do with HTML?
It's a $_GET parameter. When you submit the code, the page receiving it will be able to use $_GET['returnto'] to return you to the page you're currently on.
Take some time to learn about $_GET

Php help find where is syntax error [closed]

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Closed 8 years ago.
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foreach($db->fetch_array("SELECT id_categories FROM csn_categories_join_kartes where id_kartes===".$card['id']."") as $kat){
echo (kat['id_categories']);
}
table cols and values are all matched, something is wrong in this part of code
I tried adding $ before kat and using only one "=", sill doesnt work
NEW LINK
http://pastebin.com/RPK7vEaJ
this
where id_kartes===".$card['id']."
would be
where id_kartes=".$card['id']."
and missing $
echo $kat['id_categories'];
so full code :-
foreach($db->fetch_array("SELECT id_categories FROM csn_categories_join_kartes where id_kartes='".$card['id']."'") as $kat){
echo $kat['id_categories'];
}
best practice if you store your query result in a variable and loop over this variable.
foreach($db->fetch_array("SELECT id_categories FROM csn_categories_join_kartes where id_kartes=".$card['id']."") as $kat)

PHP error for login session [closed]

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Closed 8 years ago.
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I have an error in a single line of code but I can't seem to find it. I've tried changing the quotes to all match but that doesn't do anything. Please help. here is where the error is
$managerID = preg_replace('#[^0-9]#i',",$_SESSION["id"]);
try this :
$managerID = preg_replace('#[^0-9]#i','',$_SESSION["id"]);
btw i dont know what you are trying to do i just solved your syntax error ;)

PHP - Using $_POST after a link [closed]

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Closed 8 years ago.
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I have a form with a textbox with name="name".
In my code, I use a direct image hosted on another website in the format:
$grav_url = "http://yourwebsitehere.com/avatarimage.php?username=YourUsername";
I want the YourUsername part to be replaced with the input of the textbox.
For this to work, i'm trying the following code:
$grav_url = "http://yourwebsitehere.com/avatarimage.php?username="+ $_POST['name'];
What am I doing wrong? PHP noob here
in php . is used for concatenate string not +
here try this
$grav_url = "http://yourwebsitehere.com/avatarimage.php?username=".$_POST['name'];

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