如何成功显示记录？

Question

<?php
        $id=$_SESSION['id'];
        //$date1=date("Y-m-d");
        //echo $date1;
        $work=$_POST['postinput0'];
            //echo $work;
        $project_name=$_POST['postinput1'];
        $skills=$_POST['postinput2'];
        $description=$_POST['postinput3'];
        $file1=$_POST['file1'];
        $budget=$_POST['postinput4'];
        //$radio=$_POST['radio'];
        $sql = "INSERT INTO Job_post
                (u_id,job_category,project_name,job_sub_category,project_description,image,budget) 
                 VALUES('$id','$work','$project_name','$skills', '$description','$file1','$budget')";
            if(mysql_query( $sql, $conn ))
            {
            echo "successfully";
            }
else
{
echo "not successfully inserted";
}




    ?>

show message before submit button. that i dont want> plz help me.

show message after submit button...... is any error in my code

Answer 1

Use this

<?php
    $id=$_SESSION['id'];
    //$date1=date("Y-m-d");
    //echo $date1;
    $work=$_POST['postinput0'];
    //echo $work;
    $project_name=$_POST['postinput1'];
    $skills=$_POST['postinput2'];
    $description=$_POST['postinput3'];
    $file1=$_POST['file1'];
    $budget=$_POST['postinput4'];
    //$radio=$_POST['radio'];
    $sql = "INSERT INTO Job_post (u_id,job_category,project_name,job_sub_category,project_description,image,budget) VALUES('$id','$work','$project_name','$skills', '$description','$file1','$budget')";
    $query = mysql_query( $sql, $conn );
    if(isset($query))
    {
        echo "successfully";
    }
    else
    {
        echo "not successfully inserted";
    }
?>

Edit 01

use isset($_POST[''])

<?php
    if(isset($_POST['submit']))
    {
        $id=$_SESSION['id'];
        //$date1=date("Y-m-d");
        //echo $date1;
        $work=$_POST['postinput0'];
        //echo $work;
        $project_name=$_POST['postinput1'];
        $skills=$_POST['postinput2'];
        $description=$_POST['postinput3'];
        $file1=$_POST['file1'];
        $budget=$_POST['postinput4'];
        //$radio=$_POST['radio'];
        $sql = "INSERT INTO Job_post (u_id,job_category,project_name,job_sub_category,project_description,image,budget) VALUES('$id','$work','$project_name','$skills', '$description','$file1','$budget')";
        $query = mysql_query( $sql, $conn );
        if(isset($query))
        {
            echo "successfully";
        }
        else
        {
            echo "not successfully inserted";
        }
    }
?>