Mysql Select

[forumnz]

Please help - I am realy having trouble with this.

 

It comes up with error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /var/www/html/show_mid.php on line 12

 

Code:

<?php
$con = mysql_connect("localhost","aaa","aaa");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("photomagik_co_nz_-_soho", $con);

$result = mysql_query("SELECT * FROM show");

while($row = mysql_fetch_array($result))
  {
  echo $row['title'] . " " . $row['link'];
  echo "<br />";
  }

mysql_close($con);
?>

Thanks,

Sam.

[trq]

Your not far wrong, but you should always do some error handling when dealing with a database. Try...

 

<?php

  if (!$con = mysql_connect("localhost","aaa","aaa")) {
    die('Could not connect: ' . mysql_error());
  }

  mysql_select_db("photomagik_co_nz_-_soho", $con);

  $sql = "SELECT * FROM show";
  if ($result = mysql_query($sql)) {
    if (mysql_num_rows($result)) {
      while ($row = mysql_fetch_array($result)) {
        echo $row['title'] . " " . $row['link'] . "<br />";
      }
    } else {
      echo "No records found<br />";
    }
  } else {
    echo "Query failed<br />$sql<br />" . mysql_error();
  }

  mysql_close($con);

?>

 

What do you get now?

[forumnz]

Thanks,

 

Now I get

Query failed
SELECT * FROM show
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'show' at line 1

 

What now?

Sam.

[Barand]

SHOW is an SQL reserved word.

 

Change name of you table or use "SELECT * FROM `show` "

[beboo002]

in my machine ur code is  working without any error

 

check this

 

<?php

$con = mysql_connect("localhost","root","");

if (!$con)

  {

  die('Could not connect: ' . mysql_error());

  }

 

mysql_select_db("timesheet", $con);

 

$result = mysql_query("SELECT * FROM test1");

 

while($row = mysql_fetch_array($result))

  {

  echo $row['firstname'] . " " . $row['age'];

  echo "<br />";

  }

 

mysql_close($con);

?>

 

result is

 

tom  25

[Barand]

No, your code is working on your machine, not his code.

 

Your table name "test1" isn't an SQL reserved word.

[beboo002]

"show" is mysql keyword is not working u hav to change tablename

[Barand]

Well done, beeboo - did you come up with that all by yourself?

 

And as you'll see in my earlier post, you can also use `show`.

[micmania1]

I dont know if this will work but you don't get anywhere in life if you don't try...

 

Change...

<?php
while($row = mysql_fetch_array($result))
?>

 

to...

<?php
while($row = mysql_fetch_array($result, MYSQL_ASSOC))
?>

 

 

[beboo002]

both code will producee same result

wht are diffrance between them

[wildteen88]

Oh please... will you stop posting code suggestions. Its not do with the code. Its to do with the SQL query being used. All is explained in Barand's post above.

[Barand]

beeboo,

 

One will return an array indexed by both field name and field number, the second will return an array indexed on field name only.

[Barand]

wildteen,

 

I agree. I was expecting someone to suggest a change of background color in the stylesheet next