PHP form Echo change format [closed] - php

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I have a mobile number
0827910119
this is how it is saved in SQL, if I echo it back on a form i would like to have it shown as follow
27827910119
This is my echo command
<input type="text" name="member_msisdn" value="<? echo $rows['member_msisdn']; ?>"
readonly id="member_msisdn" />
I have tried a Session command to drop the 0 and add the 27 in front but not getting it to work
$_SESSION['msisdn'] = "27" . substr($rows['member_msisdn'], 1);
Is there another way I can echo the number to drop the 0 and add the 27 as I need the 27827910119 in my action PHP to work and not the 0827910119 number


Use
$rows['member_msisdn'] = "27" . substr($rows['member_msisdn'], 1);
instead of
$_SESSION['msisdn'] = "27" . substr($rows['member_msisdn'], 1);
or print using
<? echo $_SESSION['msisdn']; ?>

Related

Assign empty string to php variable [closed]

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Closed 9 years ago.
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Just need help fixing this php code
if ($cmtx_set_name_value == "anonymous")
{
$cmtx_set_name_value = ""
}
thanks

Try adding ; after the assignment.
$cmtx_set_name_value = "";

your code is work fine just put ; to end of line
if ($cmtx_set_name_value == "anonymous"){
$cmtx_set_name_value = "" ; }

preg match for known value plus 7 charecters [closed]

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Can anyone help me get the value of IFC/USD 0.00001031 from the webpage http://infinitecoin.com.
I need to grab the value after IFC/USD..
The way I thought to do it was to:
$file_string = file_get_contents('http://infinitecoin.com');
preg_match('/IFC/USD plus next 11 charecters', $file_string, $title);
$value = $title[1];
echo $title_out ;
But im not sure how to ask PHP to find IFC/USD then return the next 11 characters after that.
If I could accomplish this, my task would be solved..
Any help would be great.
Thank you
Jason

$output = array();
preg_match("/(?<=IFC\/USD\s)[\d\.]+/", 'IFC/USD 0.00001015', $output);
$output will look like this:
Array
(
[0] => 0.00001015
)

Parse error: syntax error, unexpected T_VARIABLE on line 5 [closed]

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I am new to PHP and am encountering this parsing error. My code is:
<?php
$con=mysqli_connect("glcmcrusaders.ipagemysql.com", "cycregister", "Crusaders13!",
"registration")
$sql="INSERT INTO camperregistration (camperFirstName, camperLastName, camperGender,
camperBirthdate, camperStreet, camperCity, camperState, camperZip, camperCountry,
camperShirtSize, guardianFirstName, guardianLastName, guardianStreet, guardianCity,
guardianState, guardianZip, guardianCountry, guardianCell, guardianEmail,
alternateFirstName, alternateLastName, alternateStreet, alternateCity, alternateState,
alternateZip, alternateCountry, alternateCell, alternateRelationship, camperSwimming)
VALUES
('$_POST[camperFirstName]', '$_POST[camperLastName]', '$_POST[camperGender]',
'$_POST[camperBirthdate]', '$_POST[camperStreet]', '$_POST[camperCity]',
'$_POST[camperState]', '$_POST[camperZip]', '$_POST[camperCountry]',
'$_POST[camperShirtSize]', '$_POST[guardianFirstName]', '$_POST[guardianLastName]',
'$_POST[guardianStreet]', '$_POST[guardianZip]', '$_POST[guardianCountry]',
'$_POST[guardianCell]', '$_POST[guardianEmail]', '$_POST[alternateFirstName]',
'$_POST[alternateLastName]', '$_POST[alternateStreet]', '$_POST[alternateCity]',
'$_POST[alternateState]', '$_POST[alternateZip]', '$_POST[alternateCountry]',
'$_POST[alternateCell]', '$_POST[alternateRelationship]', '$_POST[camperSwimming]')";
mysqli_close($con);
?>
Line 5 starts at the %sql ="INSERT INTO... ends at camperSwimming)
I am not sure what I am doing wrong.

Missing a semi-colon:
$con=mysqli_connect("glcmcrusaders.ipagemysql.com", "cycregister", "Crusaders13!",
"registration") // <-- HERE

you need a terminator(;) at the end of $con.
$con=mysqli_connect("glcmcrusaders.ipagemysql.com", "cycregister", "Crusaders13!",
"registration");
^Here

Strange call to a member function on a non-object php [closed]

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So I create a Supermarket class and initiate two objects like this:
$items[0] = new Supermarket("Item1", 2.70);
$items[1] = new Supermarket("Item2", 1.0);
Then I call the showItem() method on the first object of the array and it works:
$items[0]->showItem();
But when I try to use a for or a foreach loop through the array to show all the items I get the non-object error. The following won't work either:
$i = 0;
$items[i]->showItem();
Any ideas?

You are not using the variable sign $
$i = 0;
$items[$i]->showItem();

change like
$i = 0;
$items[i]->showItem();
$i = 0;
$items[$i]->showItem();

$items[$i] instead of $items[i]

$variable = $url + static_page.php [closed]

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I would like to uplaod files using cURL but can't figure out how I can use $url array.
For example:
$urls = array("http://images.domain.com/",
"http://flash.domain.com/",
"http://other.domain.com/"
);
foreach ($urls as $url) {
I'm trying this but without success:
$upload = "$url ."upload/upload.php";
Any advice? :)
Thanks

You need to take out the first quote for it to work
$urls = array("http://images.domain.com/",
"http://flash.domain.com/",
"http://other.domain.com/"
);
foreach ($urls as $url) {
$upload = $url ."upload/upload.php";
}

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