How to make visible of unvisible error while inserting data into mariadb database?

[eaglehopes]

Hi all, 

I have a php code which did not give any error(I think) but nothing inserted into the database.  I am trying to insert all file structure and contents of the files into database by getting info from them and supply these into to insertData function:  

<?php 
	ini_set('display_errors', '1');
	error_reporting(E_ALL);
	//mysqli_report(MYSQLI_REPORT_ALL); // can give a fatal errors ans stops php script!
	mysqli_report(MYSQLI_REPORT_OFF);

function insertData($folderName,$subFolder,$file,$parentId,$dateAdded,$dateEdited,$summary,$title,$content) {
$subCategoryIdSQLQuery= "SELECT id FROM `subCategory` WHERE  subCategory.path = '/pages/blog/$folderName/'"; 			
$con=OpenConnection(); // < OPEN DATABASE CONNECTION
$sql2 = mysqli_prepare($con,$subCategoryIdSQLQuery); // < SQL Statement 1 
  if($sql2 !== FALSE ) {  
    $result2 = executeSQLStatement($con,$subCategoryIdSQLQuery);
    $subcategoryId= $result2->fetch_row()[0]; 
    $pathy="/pages/blog/".$folderName."/".$subFolder."/".$file;
    $fileInsertQuery="INSERT INTO `files`  
                     (path,categoryId,subCategoryId,dateAdded,dateEdited,title,summary,content,active) 
                     VALUES(
                     '$pathy',
                     '$parentId',
                     '$subcategoryId',
                     '$dateAdded',
                     '$dateEdited',
                     '$title',
                     '$summary',
                     '$content',
                     '1')";
      $sql3=mysqli_prepare($con,$fileInsertQuery); // < SQL STATEMENT 2
        if( $sql3 !== FALSE ) {
          $result3 = executeSQLStatement($con,$fileInsertQuery);
        } else {
          echo mysqli_error($con)." at file : ".$folderName."/".$subFolder."/".$file; // for sql3
        }
    } else {
      echo mysqli_error($con); // for sql2
    }
    closeConnection($con); // < CLOSE DATABASE CONNECTION
  
}

function CloseConnection($conn) {
		 mysqli_close($conn);
	}
	
function executeSQLStatement($con,$query) {
    if($con) {
      return $con->query($query); 
    } else {
      return null; 
    }
}  

function OpenConnection() {
	  $dbhost ="hostname";
  	  $dbuser ="username";
      $dbpass ="password";
      $db = "filesDb";
      global $conn; // DEFINE  conn as global variable to be able to use in other functions as well!
      $conn = new mysqli($dbhost, $dbuser, $dbpass,$db); 
      if( $conn == true ) { 
        //printf("Success... %s\n", $conn->host_info);
      } else if ($conn->connect_errno) { 

        //printf("Connection failed!... %s\n", $conn->host_info);
        throw new RuntimeException('mysqli connection error: ' . $conn->connect_error);
        exit("Connection failed: %s\n".$conn->connect_error) ; 
      } 
      return $conn;
}
  
  
?>

I could not insert any info into database! Where did I do wrong? Is there any way to make error(if there is any) visible? Thanks.

[Barand]

15 minutes ago, eaglehopes said:

Is there any way to make error(if there is any) visible?

One way is not to turn off the error reporting...

 

16 minutes ago, eaglehopes said:

mysqli_report(MYSQLI_REPORT_OFF);

 

[eaglehopes]

Thanks Barand, I opened error reporting and it gives :

Fatal error: Uncaught mysqli_sql_exception: Duplicate entry '/fold1/' for key 'PATH' in /htdocs/filer.php:728

What does that mean? I am looking the table and see no duplicate entry! How can I prevent duplicate entry ?

[Barand]

4 minutes ago, eaglehopes said:

I am looking the table and see no duplicate entry!

You won't if trying to enter a duplicate is throwing an error.

Why is "path" the primary key?

6 minutes ago, eaglehopes said:

How can I prevent duplicate entry ?

Do you ...

• want to prevent duplicate entries (which it is doing now)?  or
• do you want to allow them and stop it reporting them? or
• do you want to prevent them but not report them if they are attempted?

[eaglehopes]

third option : "I want to prevent them but not report them if they are attempted?".PATH is not the primary but the unique key, because it is unique for each subcategory. 

[Barand]

Catch the error. If it's a dupe key error (1062) then continue, otherwise throw the error

try {
   insert a new row
}
catch (error) {
   if error number != 1062 {         // if it's not a dupe error
       throw error                   // report it.
   }
}

 

[eaglehopes]

Thanks Barand, I did and I saw other errors at last and finally I noticed that the error is in dateTime format! Thanks. I think I can solve it now! My code was :

try {
  $result1 = executeSQLStatement($con,$subCategoryInsertQuery);
}
  catch (Exception $e) {
  if( mysqli_errno($con) != 1062 ) {         // if it's not a dupe error
  	echo $e ;                   // report it.
  } 
}

 

[Barand]

You should

throw $e;

not

echo $e;

 

[eaglehopes]

Ok.