PHP: Executing Select statement

Question

I have entered the following query in MySQL:

insert into hospital (name,age) values ('william', 'select * from department where age = $agegrp');

Now I have fetched this value (the sql), and trying to execute it. How can i do this?

I have fetched the select statement into a variable called $var.

$agegrp = "10";
$value = mysql_query ($var) or die ('error');

I only get error so something is wrong with my query that I inserted. How can I solve this?

It would be better to have die( mysql_error() ) and then you could see what the error is (I won't spoil the surprise). — Pete
– Pete, Commented Jul 13, 2012 at 11:54
where and how do you set $var? could you show your complete code if the mysql_error() doesn't help you? — Nanne
– Nanne, Commented Jul 13, 2012 at 11:55
Please don't use mysql_* functions in new code. They were removed from PHP 7.0.0 in 2015. Instead, use prepared statements via PDO or MySQLi. See Why shouldn't I use mysql_* functions in PHP? for more information. — Madara's Ghost
– Madara's Ghost, Commented Jul 13, 2012 at 12:01

Jan Prieser · Accepted Answer · 2012-07-13 12:00:07Z

1

first question for me: why do you store sql code in your hospital table? it would make more sense if all your sql code would be embedded in your php program code. if you later change your table design, you won't have to change the content of the database table.

but to answer your question, the content of your $var is a simple string. what you want is to evaluate it or to replace the string '$agegrp' with the actual value of the variable $agegrp. so you could do:

$agegrp = "10";
$var = str_replace('$agegrp',$agegrp,$var;
$value = mysql_query ($var) or die ('error');

this would be a simple solution.

answered Jul 13, 2012 at 12:00

Jan Prieser

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Comments

tbddeveloper · Accepted Answer · 2012-07-13 11:56:17Z

0

age is a property that will probably be an integer. You can't insert a query into that field. What you need to do is execute query;

"SELECT * FROM Department WHERE age = {$agegrp}"

Then, get the result of that and then execute your insert. Also, look at bind parameters for your queries early on. You don't want to allow sql injection.

answered Jul 13, 2012 at 11:56

tbddeveloper

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1 Comment

user1523434 Over a year ago

no, the value $agegrp doesn't get replaced with 10. It doesn't get executed. when i echo it just prints "SELECT * FROM Department WHERE age = {$agegrp}"

Mushahid Hussain · Accepted Answer · 2012-07-13 12:01:54Z

0

Try this one.

 <?php
    $con = mysqli_connect("localhost","username","password");
    if (!$con)
      {
      die('Could not connect: ' . mysqli_error());
      }

  $var=  mysqli_select_db("my_db", $con);

    $var="select * from department where age = $agegrp");
    $result=mysqli_query($var);
    while($row = mysqli_fetch_array($result))
      {
      echo $row['age'] ;
      echo "<br />";
      }

mysqli_close($con);
?>

answered Jul 13, 2012 at 12:01

Mushahid Hussain

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1 Comment

Madara's Ghost Over a year ago

Please don't use mysql_* functions in new code. They were removed from PHP 7.0.0 in 2015. Instead, use prepared statements via PDO or MySQLi. See Why shouldn't I use mysql_* functions in PHP? for more information.

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