php code into variable

Question

I want to insert a PHP code into a variable.

I want to replace "Some text" with a PHP code.

I found some examples on the internet, but when I try to run my script it isn't working.

This is my current script: $purchaseID = "Some text";

I want to change "Some text" with the following code:

<?php
  echo date('y/m/d');
  echo "O";
  $link = mysql_connect("localhost", "root", "");
  mysql_select_db("db", $link); 
  $result = mysql_query("SELECT * FROM table", $link);
  $num_rows = mysql_num_rows($result);echo "$num_rows"+1;
?>

I have tried:

$purchaseID = <<<END
echo date('y/m/d');
echo "O"; 
$link = mysql_connect("localhost", "root", "");
mysql_select_db("db", $link);
$result = mysql_query("SELECT * FROM table", $link);
$num_rows = mysql_num_rows($result);
echo "$num_rows"+1;
END;

When I run this I get a error.

How can I fix this and replace "Some text" with the PHP code?

Why do you want to have "some code into a variable", what you expect is to have $purchaseID: "2016/06/24O5" (if $num_rows = 4) is that right? — olibiaz
– olibiaz, Commented Jun 25, 2016 at 2:18

olibiaz · Accepted Answer · 2016-06-25 02:25:43Z

0

If i understand well your need, you just need to concatenate the different part of the string:

$purchaseID = date('y/m/d');
$purchaseID .= "O"; 
$link = mysql_connect("localhost", "root", "");
mysql_select_db("db", $link);
$result = mysql_query("SELECT * FROM table", $link);
$num_rows = mysql_num_rows($result);
$purchaseID .= $num_rows + 1;

You will get (considering $num_rows to be 4)

2016/06/24O5

Do you really need the heredoc syntax here?

answered Jun 25, 2016 at 2:25

olibiaz

2,6034 gold badges31 silver badges33 bronze badges

Sign up to request clarification or add additional context in comments.

5 Comments

the_boy_za Over a year ago

I get a error message. When I vardump $purchaseID ill get the right string. When I vardump $requestTransaction ($purchaseID is a part of this) I get: "Transaction.purchaseId does not match format."

olibiaz Over a year ago

Edit your question to explain what is $requestTransaction and what do you mean by $purchaseID is part of this. But this is another problem right? Does your initial problem is fixed?

the_boy_za Over a year ago

I dont know if my problem is fixed. The script is still not working by some reason. It is also possible that the script is not working because the solution I tried is not working

olibiaz Over a year ago

We can't really help you since we don't know nothing about $requestTransaction

Pinke Helga Over a year ago

No "snake questions" please. Let new aspects like $requestTransaction go into a completely new question.

Pinke Helga · Accepted Answer · 2016-06-25 02:27:11Z

0

You are using the heredoc syntax, that interpolates variables. You might want the herenow syntax without interpolation:

$purchaseID = <<<'END'
    echo $variable;
END;

// $purchaseID == "echo $variable;"

Enclose the delimiter into single qoutes. Interpolation is not done in this case. To execute the code, you have to use eval($purchaseID);. Be aware, that eval has a dangerous potential when relying on external data.

edited Jun 25, 2016 at 2:27

answered Jun 25, 2016 at 2:20

Pinke Helga

6,7622 gold badges25 silver badges46 bronze badges

2 Comments

the_boy_za Over a year ago

I get a error message. When I vardump $purchaseID ill get the right string. When I vardump $requestTransaction ($purchaseID is a part of this) I get: "Transaction.purchaseId does not match format."

Pinke Helga Over a year ago

You should create a new question targeting the problem to form a proper transaction format. Provide all the necessary code with the new question. You can simplify the example code by hardcoding the expected value into $purchaseID

Community · Accepted Answer · 2017-05-23 10:34:05Z

0

try but i don't recommend you eval

There are some side effects using eval() refer and learn here

<?php


$time =  '<?php echo time(); ?>'; 

eval('?>'.$time.'<?php;');


?>

edited May 23, 2017 at 10:34

CommunityBot

11 silver badge

answered Jun 25, 2016 at 4:08

JYoThI

12.1k1 gold badge13 silver badges26 bronze badges

Comments

Ikari · Accepted Answer · 2016-06-25 04:57:13Z

0

You would have to escape the variables, like this:

$purchaseID = <<<END
echo date('y/m/d');
echo "O"; 
\$link = mysql_connect("localhost", "root", "");
mysql_select_db("db", \$link);
\$result = mysql_query("SELECT * FROM table", \$link);
\$num_rows = mysql_num_rows(\$result);
echo "\$num_rows"+1;
END;

That's it!

Cheers!

answered Jun 25, 2016 at 4:57

Ikari

3,2443 gold badges31 silver badges35 bronze badges

Collectives™ on Stack Overflow

php code into variable

4 Answers 4

5 Comments

2 Comments

Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

5 Comments

2 Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related