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mysqli query Not returning rows but error of mysql_query() expects parameter 2 to be resource

Posted on 2011-09-06
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Last Modified: 2012-05-12
I worked with this problem for about 3 hours.  So-called soluiions do Not work and some are mixed mysqli  and mysql.  Below is some code that gives the error of "mysql_query() expects parameter 2 to be resource".  I have no idea what resource they are talking about!

$conn1 = new mysqli($hostName1, $userName2, $password2, $databaseName1) ;
/* check connection */
      if (mysqli_connect_errno()) {
            printf("Connect failed: %s\n", mysqli_connect_error());
            exit(); // <==///========
      }
    echo ' database ' . $databaseName1 . ' is now connected to program. ' ;
   
      // Create SQL code to insert a record into the Prospect table.
      $SQL1 = "INSERT INTO Prospect (CustomerYN, ActiveYN, FirstName, LastName, EmailMain, CoProfile) ";
      $SQL1 .= " VALUES('N', 'Y', 'John', 'Hammer', 'John@HammondTrees.com', 'Test only in 2011' )";
          
   $conn1->query($SQL1); // run the query
      printf (" |  New Record has id %d.\n", $conn1->insert_id);
      echo ' \  Prospect table ' ;
// this works OK, but not the Select section below
      
      // use a Select to show current records
      $SQL1 = "SELECT CustomerYN, ActiveYN, FirstName, LastName, EmailMain, CoProfile " ;
      $SQL1 = " FROM Prospect " ;
      // $result1 = $conn1->query($SQL1) ;
      // $result1 = mysql_query( $link1, $SQL1) ;
      // associative array
      // $row22 = mysqli_fetch_array($result1, MYSQLI_ASSOC);
      
       echo "CustomerYN, ActiveYN, FirstName, LastName, EmailMain, CoProfile " ;
       
       // $last1  = mysql_query(   $SQL1 , $link1) ;
      $last1  = mysql_query(   $SQL1  ) ;
       $totalRows1 = mysql_num_rows($last1);
       while ( $addToArray1 = mysql_fetch_array($last1) )
       {
                // This could be placed in a table.
                echo " " . $row['CustomerYN'] . "  " ;
                echo "<br />" ;
       }
0
Comment
Question by:KennethSumerford1
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6 Comments
 
LVL 9

Expert Comment

by:jeff_01
ID: 36493780
I think the issue is here

 $SQL1 = "SELECT CustomerYN, ActiveYN, FirstName, LastName, EmailMain, CoProfile " ;
      $SQL1 = " FROM Prospect " ;


TRY


 $SQL1 = "SELECT CustomerYN, ActiveYN, FirstName, LastName, EmailMain, CoProfile " ;
 $SQL1 .= " FROM Prospect " ;

0
 
LVL 83

Expert Comment

by:Dave Baldwin
ID: 36494004
You can't mix 'mysql' and 'mysqli' functions and expect thing to work right.  This page http://us2.php.net/manual/en/mysqli.query.php shows two different ways to use 'mysqli' and this page http://us2.php.net/manual/en/function.mysql-query.php shows a similar thing for 'mysql' .  If you try to mix the two different methods, your results will be unpredictable.
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LVL 83

Accepted Solution

by:
Dave Baldwin earned 250 total points
ID: 36494079
I think this will work for you.  There are at least two other ways of doing this.
$conn1 = new mysqli($hostName1, $userName2, $password2, $databaseName1) ;
/* check connection */
if (mysqli_connect_errno()) {
	printf("Connect failed: %s\n", mysqli_connect_error());
	exit(); // <==///========
	}
echo ' database ' . $databaseName1 . ' is now connected to program. ' ;
   
// Create SQL code to insert a record into the Prospect table.
$SQL1 = "INSERT INTO Prospect (CustomerYN, ActiveYN, FirstName, LastName, EmailMain, CoProfile) ";
$SQL1 .= " VALUES('N', 'Y', 'John', 'Hammer', 'John@HammondTrees.com', 'Test only in 2011' )";
          
$conn1->query($SQL1); // run the query
printf (" |  New Record has id %d.\n", $conn1->insert_id);
echo ' \  Prospect table ' ;
// this works OK, but not the Select section below
      
// use a Select to show current records
$SQL2 = "SELECT CustomerYN, ActiveYN, FirstName, LastName, EmailMain, CoProfile " ;
$SQL2 .= " FROM Prospect " ;
$result2 = $conn1->query($SQL2) ;
while($row22 = $result2->fetch_array(MYSQLI_ASSOC))) {
      
echo "CustomerYN = ".$row22["CustomerYN"].", ActiveYN = ".$row22["ActiveYN"].", ";
echo "FirstName = ".$row22["FirstName"].", LastName = ".$row22["LastName"].", "
echo "EmailMain = ".$row22["EmailMain"].", CoProfile = ".$row22["CoProfile"]."<br>";
}

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LVL 9

Expert Comment

by:jeff_01
ID: 36494219
Either way the query will not work with the "." missing as effectively the query would be just  

mysql_query("FROM prospect");

instead of

mysql_query("SELECT CustomerYN, ActiveYN, FirstName, LastName, EmailMain, CoProfile FROM prospect");



0
 

Author Closing Comment

by:KennethSumerford1
ID: 36497701
I had to add two ;'s and then it worked great!  Thanks soooo much!  The code was accurate, complete and easy for a beginner PHP programmer to understand.  I have more than 10 years experience with VB but PHP has some weird coding, at times.  I did take your code, make a few edits and create the function below.

function fnAddRecSeletRecs($hostName1, $userName3, $password3, $databaseName3 ) {
            // Add one record and display all records in the Prospect table.
            
      try {
            $conn1 = new mysqli($hostName1, $userName3, $password3, $databaseName3) ;
      /* check connection */
      if (mysqli_connect_errno()) {
            printf("Connect failed: %s\n", mysqli_connect_error());
            exit(); // <==///========
            }
      echo ' database ' . $databaseName3 . ' is now connected to program. ' ;
        
      // Create SQL code to insert a record into the Prospect table.
      $SQL1 = "INSERT INTO Prospect (CustomerYN, ActiveYN, FirstName, LastName, EmailMain, CoProfile) ";
      $SQL1 .= " VALUES('N', 'Y', 'John', 'Hammer', 'John@HammondTrees.com', 'Test only in 2011' )";
                
      $conn1->query($SQL1); // run the query
      printf (" |  New Record has id %d.\n", $conn1->insert_id);
      echo ' \  Prospect table ' ;
      // this works OK, but not the Select section below
            
      // use a Select to show current records
      $SQL2 = "SELECT CustomerYN, ActiveYN, FirstName, LastName, EmailMain, CoProfile " ;
      $SQL2 .= " FROM Prospect " ;
      $result2 = $conn1->query($SQL2) ;
      $K1 = 0 ;

      while($row22 = $result2->fetch_array(MYSQLI_ASSOC)) {
            $K1 = $K1 + 1 ;
      echo $K1 . "  ";      
      echo "CustomerYN = ".$row22["CustomerYN"].", ActiveYN = ".$row22["ActiveYN"].", ";
      echo "FirstName = ".$row22["FirstName"].", LastName = ".$row22["LastName"].", ";
      echo "EmailMain = ".$row22["EmailMain"].", CoProfile = ".$row22["CoProfile"]."<br>";
      } // == end of while
      
      /* close connection */
      $conn1->close();
      return true ; // <===///==========
 } // === end of try

 catch ( Exception $e1) {
      echo "  error in database and table operations-- " . $e1 . " | " ;
                return false ; // <===///==========
  }
 } // === end of function ======================
      
 --- Kenneth
0
 
LVL 83

Expert Comment

by:Dave Baldwin
ID: 36497903
You're welcome.  I didn't have a chance to test it since I didn't have a matching database so I'm not surprised I missed something.
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