Error in supplying arguement code

Question

Tonkz 0 Light Poster

15 Years Ago

Anyone knows what I'm doing wrong here? I am getting supplied argument is not a valid MySQL result resource in line 34

<?php
//Connect to database server
$conn=mysql_connect("mywebsite", "user", "pass");
if (!$conn)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("data", $conn);

?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>gg</title>
</head>
<body>

<?php

$result= mysql_query("SELECT student.First_Name AS first,
                          student.Last_Name AS last,
	     	          faculty.First_Name AS afirst,
                          faculty.Last_Name AS alast,
                          faculty.Office AS office,
	     	          faculty.Phone AS phone
                          FROM Student
	                  INNER JOIN Faculty
	                  ON Student.Advisor_ID=Faculty.Faculty_ID
                          ORDER BY last, first, afirst, alast, office, phone;");
	  
if (!$result) {

      die('Invalid query: ' . mysql_error());

      }

   while ($row = mysql_fetch_array($result)) {
	  
	  echo '<h2>Students and Advisors</h2>';
	  echo '<table width="100%" border="2">';
 	  echo '<tr>';
          echo '<th>Student First Name</th>';
          echo '<th>Student Last Name</th>';
          echo '<th>Student Advisors First Name</th>';
          echo '<th>Student Advisors Last Name</th>';
	  echo '<th>Advisor Office</th>';
	  echo '<th>Advisor Phone</th>';
 	  echo '</tr>';
	  echo '<tr>';
	  echo "<td>".$row['first']."</td>";
	  echo "<td>".$row['last']."</td>";
	  echo "<td>".$row['afirst']."</td>";
	  echo "<td>".$row['alast']."</td>";
	  echo "<td>".$row['office']."</td>";
	  echo "<td>".$row['phone']."</td>";
	  echo '</tr>';
	  echo '</table>';
	  }
	  mysql_close($conn);		  	  
?>
</body>
</html>

php

2 Contributors
13 Replies
194 Views
7 Hours Discussion Span
Latest Post 15 Years Ago Latest Post by Tonkz

ShawnCplus 456 Code Monkey

15 Years Ago

Your query is failing. Try placing this after your query call

if (!$result) {
    die('Invalid query: ' . mysql_error());
}

ShawnCplus 456 Code Monkey

15 Years Ago

Not necessarily the problem but this code

if (!$conn)
{
  die('Could not connect: ' . mysql_error());
}

Should be moved to line 4.

ShawnCplus 456 Code Monkey

15 Years Ago

Interesting because I don't see anywhere in your code where you're querying a 'data' table. If it couldn't find the database it would give you Unknown database 'data' not Table 'data' doesn't exist .

Is that all of the code or is that just a bit of it? Also, repost your code with the changes.

ShawnCplus 456 Code Monkey

15 Years Ago

well yeah, the fact that you left out that it said data.Student is pretty big :) lowercase Student on line 32 , case is important, that should fix it

Tonkz commented: Thank You +1

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Tonkz 0 Light Poster · Answer 1 · 2009-05-06T23:02:54+00:00

It says "Invalid query: Table 'data' doesn't exist"
Is that mean it can't find my databasE?

Tonkz 0 Light Poster · Answer 2 · 2009-05-06T23:16:48+00:00

I updated the post, the same result as before "Invalid query: Table 'data' doesn't exist

Tonkz 0 Light Poster · Answer 3 · 2009-05-06T23:28:59+00:00

This code is all of it I didnt change anything besides user, pass, data, and website.
I have rechecked my codes but I can't seem to find what I did wrong.

<?php
//Connect to database server
$conn=mysql_connect("website", "user", "pass");
if (!$conn)
  {
  die('Could not connect: ' . mysql_error());
  }
mysql_select_db("data", $conn);


?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>gg</title>
</head>
<body>

<?php

$result= mysql_query("SELECT student.First_Name AS first,
             student.Last_Name AS last,
	     	 faculty.First_Name AS afirst,
             faculty.Last_Name AS alast,
             faculty.Office AS office,
	     	 faculty.Phone AS phone
      FROM Student
	  INNER JOIN Faculty
	  ON Student.Advisor_ID=Faculty.Faculty_ID
      ORDER BY last, first, afirst, alast, office, phone;");
	  

      if (!$result) {

      die('Invalid query: ' . mysql_error());

      }
   while ($row = mysql_fetch_array($result)) {
	  
	  echo '<h2>Students and Advisors</h2>';
	  echo '<table width="100%" border="2">';
 	  echo '<tr>';
      echo '<th>Student First Name</th>';
      echo '<th>Student Last Name</th>';
      echo '<th>Student Advisors First Name</th>';
      echo '<th>Student Advisors Last Name</th>';
	  echo '<th>Advisor Office</th>';
	  echo '<th>Advisor Phone</th>';
 	  echo '</tr>';
	  echo '<tr>';
	  echo "<td>".$row['first']."</td>";
	  echo "<td>".$row['last']."</td>";
	  echo "<td>".$row['afirst']."</td>";
	  echo "<td>".$row['alast']."</td>";
	  echo "<td>".$row['office']."</td>";
	  echo "<td>".$row['phone']."</td>";
	  echo '</tr>';
	  echo '</table>';
	  }
	  mysql_close($conn);		  	  
?>
</body>
</html>

ShawnCplus 456 Code Monkey Team Colleague · Answer 4 · 2009-05-06T23:31:58+00:00

K, last addition, if this doesn't give us something then I'm fresh out of ideas. replace mysql_select_db() with

$dbselect = mysql_select_db('data', $conn);
if(!$dbselect) {
  die('Couldn\'t use database: ' . mysql_error());
}

Tonkz 0 Light Poster · Answer 5 · 2009-05-06T23:46:28+00:00

Still the same thing "Invalid query: Table 'data.Student' doesn't exist"
well I left out ".Student" before see if that helps

<?php
//Connect to database server
$conn=mysql_connect("website", "user", "pass");
if (!$conn)
  {
  die('Could not connect: ' . mysql_error());
  }
$dbselect = mysql_select_db('data', $conn);
if(!$dbselect) {
 die('Could not find data: ' . mysql_error());
}

?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Students and Advisors</title>
</head>
<body>

<?php

$result= mysql_query("SELECT student.First_Name AS first,
             student.Last_Name AS last,
	     	 faculty.First_Name AS afirst,
             faculty.Last_Name AS alast,
             faculty.Office AS office,
	     	 faculty.Phone AS phone
      FROM Student
	  INNER JOIN Faculty
	  ON Student.Advisor_ID=Faculty.Faculty_ID
      ORDER BY last, first, afirst, alast, office, phone;");
	  

      if (!$result) {

      die('Invalid query: ' . mysql_error());

      }
   while ($row = mysql_fetch_array($result)) {
	  
	  echo '<h2>Students and Advisors</h2>';
	  echo '<table width="100%" border="2">';
 	  echo '<tr>';
      echo '<th>Student First Name</th>';
      echo '<th>Student Last Name</th>';
      echo '<th>Student Advisors First Name</th>';
      echo '<th>Student Advisors Last Name</th>';
	  echo '<th>Advisor Office</th>';
	  echo '<th>Advisor Phone</th>';
 	  echo '</tr>';
	  echo '<tr>';
	  echo "<td>".$row['first']."</td>";
	  echo "<td>".$row['last']."</td>";
	  echo "<td>".$row['afirst']."</td>";
	  echo "<td>".$row['alast']."</td>";
	  echo "<td>".$row['office']."</td>";
	  echo "<td>".$row['phone']."</td>";
	  echo '</tr>';
	  echo '</table>';
	  }
	  mysql_close($conn);		  	  
?>
</body>
</html>

Tonkz 0 Light Poster · Answer 6 · 2009-05-06T23:53:19+00:00

Tonkz 0 Light Poster

15 Years Ago

hahaha, Wupee that works! :D Thank you !!

Tonkz 0 Light Poster · Answer 7 · 2009-05-07T00:12:00+00:00

One silly question how do you repeat the table ("<td></td>") as long as the statement is true because my other data wasn't inside the table

<?php
//Connect to database server
$conn=mysql_connect("website", "user", "password");
if (!$conn)
  {
  die('Could not connect: ' . mysql_error());
  }
$dbselect = mysql_select_db('data', $conn);
if(!$dbselect) {
 die('Could not find data: ' . mysql_error());
}

?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Students and Advisors</title>
</head>
<body>
<h2>Students and Advisors</h2>
<table width="100%" border="2">
<tr>
<th>Student First Name</th>
<th>Student Last Name</th>
<th>Student Advisors First Name</th>
<th>Student Advisors Last Name</th>
<th>Advisor Office</th>
<th>Advisor Phone</th>
</tr>

<?php
	  
$result= mysql_query("SELECT student.First_Name AS first,
             student.Last_Name AS last,
	     	 faculty.First_Name AS afirst,
             faculty.Last_Name AS alast,
             faculty.Office AS office,
	     	 faculty.Phone AS phone
      FROM student
	  INNER JOIN faculty
	  ON student.Advisor_ID=faculty.Faculty_ID
      ORDER BY last, first, afirst, alast, office, phone;");
	  

      if (!$result) {

      die('Invalid query: ' . mysql_error());

      }
	  	  
while ($row = mysql_fetch_array($result)) {
	  
	 
	  echo '<tr>';
	  echo "<td>".$row['first']."</td>";
	  echo "<td>".$row['last']."</td>";
	  echo "<td>".$row['afirst']."</td>";
	  echo "<td>".$row['alast']."</td>";
	  echo "<td>".$row['office']."</td>";
	  echo "<td>".$row['phone']."</td>";
	  echo '</tr>';
	  echo '</table>';
	  }
	  mysql_close($conn);		  	  
?>
</body>
</html>

ShawnCplus 456 Code Monkey Team Colleague · Answer 8 · 2009-05-07T00:15:07+00:00

move the echo '</table>'; outside the loop. or just put </table> before </body> and remove the echo altogether

Tonkz 0 Light Poster · Answer 9 · 2009-05-07T00:18:29+00:00

Tonkz 0 Light Poster

15 Years Ago

Oh right right thanks again :D