SQL Query Not Working In Loop

Question

I have a script that extracts several rows of data from an external file, I then need to upload each of these rows to a database. I put a query inside my loop but it does nothing. even with error reporting turned on I get nothing. I check the database but nothing is being uploaded.

What is the correct way to do this? Thanks in advance!

My Code:

$size = $_FILES['file']['size'];
$filename = $_FILES['file']['tmp_name']; 
$max_filesize = 100000;
  if($size > $max_filesize) //check file size
  die('File is too large');

if(file_exists($filename)){
$fp = fopen($filename, "r");
$str = fread($fp, filesize($filename)); //file text stored in variable
$br = "\n";          //search for new row
$rows = substr_count($str,$br);  //number of rows
echo "Rows: ". $rows."<br />"; 
$row = explode( "\n",$str);

 $x=0;
while($x<=$rows)
  {
  $field = $row[$x]; 
  $exp = explode("|",$field); 

    $case_number = $exp[1];
    $unknown1 = $exp[2];
    $chapter = $exp[3];
    $filed = $exp[8];
    //tons more variables, removed for readability

   $addrow = mysql_query("INSERT INTO records (case_number, wild1, chapter, filed) VALUES('".$case_number."', '".$unknown1."', '".$chapter."', '".$filed."')");
    if($addrow)  
    {  
    echo "<p style=\"font-weight:bold\">Row ".$x." uploaded</p>";   
    }else{
die(mysql_error());
}  

  $x++;
  }
fclose($fp); 
 }

You are not doing any error checking in your query. Add it. The manual shows how. php.net/manual/en/function.mysql-query.php The manual will also show you alternatives to the mysql library, which is outdated. — Pekka
– Pekka, Commented Nov 1, 2012 at 7:28
@Mr.Alien @Pekka, I have error_reporting(E_ALL); ini_set('display_errors','1'); at the top of my document. I'm getting no errors whatsoever. Nor does the echo at the bottom of the loop display. — I wrestled a bear once.
– I wrestled a bear once., Commented Nov 1, 2012 at 7:31
@Gordie use mysqli_error() for turning on sql errors, php.net/manual/en/mysqli.error.php — Mr. Alien
– Mr. Alien, Commented Nov 1, 2012 at 7:35
Because one of your database field is primary, you cannot insert duplicate value in the primary field, so you are getting this error — Mr. Alien
– Mr. Alien, Commented Nov 1, 2012 at 7:48
@Mr.Alien It really was as simple as that. I thought the field I set for primary was unique, turns out it's not. Sorry I'm an idiot. Make it an answer and I'll choose it. Free points. — I wrestled a bear once.
– I wrestled a bear once., Commented Nov 1, 2012 at 7:53

Mr. Alien · Accepted Answer · 2012-11-01 07:56:45Z

1

You have mentioned the error Duplicate entry for key 'PRIMARY' so you cannot have a duplicate value in the column which is set to PRIMARY.

answered Nov 1, 2012 at 7:56

Mr. Alien

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