Error in Inserting data's in table

Question

I have a code which inserts data's in each tables in a specific database. After i run it this error come up

"Parse error: syntax error, unexpected '}', expecting ';' in C:\xampp\htdocs\Thesis\database\insertdata.php on line 37"

I check the codes for typo error but still i don't see any error on it maybe it is cause by a looping error

This is my Code

$db_name = array('morning_section_masterfile','evening_section_masterfile','afternoon_section_masterfile');
for($y=0;$y<=2;$y++)
{
   $db = mysql_select_db($db_name[$y],$connectDatabase);
   $tables = array('Pilot_Sections','Black_Sections');
     for($a=0;$a<=1;$a++)
     {
       //variable making
        $teacher = array ('Jane','Jeff','Liezeth','Loremas','Canada');
       $Default_Lname = 'Lorems';
       $Default_Fname = 'Vierzehn';
       $x=0;
       //Adding 30 students in one section
    do
    {
        for($i=0;$i<=30;$i++)
        {
            $section_teacher = $teacher[$x];
            $student_section = 'IT70'.$x.'E-C';
            $student_Lname = str_shuffle($Default_Lname);
            $student_Fname = str_shuffle($Default_Fname);
            $table = $tables[$a];
            $insert = "INSERT INTO $table (section_Teacher, student_Lastname, student_Firstname, student_Section) VALUES 
            ('{$section_teacher}','{$student_Lname}','{$student_Fname}','{$student_section}')";
            $insertdata = mysql_query($insert,$connectDatabase);
        }
    //check the number of students in a section section and adding another section
        if($i==31)
        {
            $x++;
            $i=0;
        }
    } while($x<=4)
  }
}

you missed a semicolon in line no 36... – Ronser Oct 15 '14 at 05:58 — Ronser, Oct 15 '14 at 05:58
WOWeeeee, my comment was their answers – Ronser Oct 15 '14 at 06:01 — Ronser, Oct 15 '14 at 06:01
@Jeff Please use PDO find the link in my answer ... :) – Tushar Gupta Oct 15 '14 at 06:56 — Tushar Gupta, Oct 15 '14 at 06:56

score 0 · Accepted Answer · edited May 23 '17 at 11:59

You missed a semi colon after while

do
    {
        for($i=0;$i<=30;$i++)
        {
            $section_teacher = $teacher[$x];
            $student_section = 'IT70'.$x.'E-C';
            $student_Lname = str_shuffle($Default_Lname);
            $student_Fname = str_shuffle($Default_Fname);
            $table = $tables[$a];
            $insert = "INSERT INTO $table (section_Teacher, student_Lastname, student_Firstname, student_Section) VALUES 
            ('{$section_teacher}','{$student_Lname}','{$student_Fname}','{$student_section}')";
            $insertdata = mysql_query($insert,$connectDatabase);
        }
    //check the number of students in a section section and adding another section
        if($i==31)
        {
            $x++;
            $i=0;
        }
    } while($x<=4);

Also as an advice don't use mysql_* as they are deprecated, Use PDO instead to prevent sql injection. Please refer to my answer here I've explained it in detail there.

score 0 · Answer 2 · answered Oct 15 '14 at 06:00

try this, you are missing semicolon in while loop at end.

<?php
$db_name = array('morning_section_masterfile','evening_section_masterfile','afternoon_section_masterfile');
for($y=0;$y<=2;$y++)
{
   $db = mysql_select_db($db_name[$y],$connectDatabase);
   $tables = array('Pilot_Sections','Black_Sections');
     for($a=0;$a<=1;$a++)
     {
       //variable making
        $teacher = array ('Jane','Jeff','Liezeth','Loremas','Canada');
       $Default_Lname = 'Lorems';
       $Default_Fname = 'Vierzehn';
       $x=0;
       //Adding 30 students in one section
    do
    {
        for($i=0;$i<=30;$i++)
        {
            $section_teacher = $teacher[$x];
            $student_section = 'IT70'.$x.'E-C';
            $student_Lname = str_shuffle($Default_Lname);
            $student_Fname = str_shuffle($Default_Fname);
            $table = $tables[$a];
            $insert = "INSERT INTO $table (section_Teacher, student_Lastname, student_Firstname, student_Section) VALUES
            ('{$section_teacher}','{$student_Lname}','{$student_Fname}','{$student_section}')";
            $insertdata = mysql_query($insert,$connectDatabase);
        }
    //check the number of students in a section section and adding another section
        if($i==31)
        {
            $x++;
            $i=0;
        }
    } while($x<=4);
  }
}

score 0 · Answer 3 · edited Jan 07 '16 at 19:29

I this mysql is old, you can change it to mysqli

You can use this:

<?php
//******  MySql Connect
$mysqli = new mysqli("localhost", "user", "password", "database");
if ($mysqli->connect_errno) {
    echo "Connect failed";

    // if connection error script stop working
    exit();
    die();
}

//****** real_escape_string (Security)
$city = $mysqli->real_escape_string($city);

//****** RESULT
$query = "SQL";
if ($result = $mysqli->query($query)) {
    while ($row = $result->fetch_assoc()) {
       echo $row["Name"];
    }
    $result->free();
}

//****** Query
$mysqli->query("SQL");

//****** Close
$mysqli->close();

?>

Error in Inserting data's in table

3 Answers3