[SOLVED] PHP code to select certain records and display only certain fields

[jlgray48]

This program is suppose to select all records from table yahoo with id = 44 and display the field "name" only.

 

Any help is greatly appreciated.

 

 

<?php

 

$conn=mysql_connect("127.0.0.1", "odbc", "") ;

mysql_select_db("uah",$conn);

$sql = "Select name from yahoo where  id=44";

mysql_query($sql,$conn) ;

$result = mysql_query($sql,$conn);

while ($array = mysql_fetch_array($result)) {

  print "$array['name']";

};

 

?>

 

[unkwntech]

1st: What output does you code provide to you?

2nd(and nowhere near as important):

When you type out SQL syntax (in your code or not) you should allways capitalise the SQL keywords so that it is easyer to read so this:

$sql = "Select name from yahoo where  id=44";

would look like this:

$sql = "SELECT name FROM yahoo WHERE id=44"

[jlgray48]

I'm getting no output at the present.

[Trium918]

Would that be the name of a column in the database?

 

[jlgray48]

Would what be the name of a column? Name and ID are my fields.

[Trium918]

Ok what is going on. Are you getting an error or what?

 

Try this:

<?php
$conn=mysql_connect("127.0.0.1", "odbc", "") ; 
mysql_select_db("uah",$conn); 

$sql = "Select name from yahoo where  id=44";

$result = mysql_query($sql,$conn) or die("Error".mysql_error()); 

while ($array = mysql_fetch_array($result)) {
  print "$array['name']";
}
?>

[jlgray48]

No errors just blank page

[Trium918]

No errors just blank page

 

Try what I just posted!

[trq]

At the very least, always check your queries succeed before trying to use any results.

 

<?php

  $conn = mysql_connect("127.0.0.1", "odbc", "") ;
  mysql_select_db("uah",$conn);
  $sql = "SELECT name FROM yahoo WHERE  id = 44";
  if ($result = mysql_query($sql,$conn)) {
    if (mysql_num_rows($result)) {
      while ($row = mysql_fetch_array($result)) {
        print $row['name'];
      }
    } else {
      echo "No results found";
    }
  } else {
    echo "Query failed<br />$sql<br />" . mysql_error();
  }

?>

[jlgray48]

That works, can we take out all the debugging and make it simple? Do you have to have the if or is that for the debugging?

 

Thanks for all your help.

 

 

jg