Hi friends I am trying to isert this data to mysql database.

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Hi friends I am trying to isert this data to mysql database. However I cannot get this done because I do not know what is happening.

What I have tried:

HTML

<!doctype html>


<meta charset="utf-8">
<title>Untitled Document

     
  <div id="wrapper">
       
        <div class="box-header">
              <h2>NEW USER</h2>
         </div>
           
           
           
           information
           
            
               
                <div class="input-group">
                  <div class="input-group-addon">
                    </div>
                    
		  </div>
                
                  <div class="input-group">
                  <div class="input-group-addon">
                    ^__i class="fa fa-flag">
                  </div>
                    
			Albania
			Algeria
                    
		  </div>
                
                <div class="input-group space">
                  <div class="input-group-addon">
                    ^__i class="fa-building-o">
                  </div>
                    
		  </div>
				
		<div class="input-group space">
                    
				</div>
          
          		<div class="input-group space">
                  <div class="input-group-addon">
                    ^__i>AGE
                  </div>
                    
		 
                  Year(s)
                  Month(s)
                  Day(s)
                 
		</div>
          
                  <div class="form-group space">
                   Gender:
                  
                   Male
                    
                    Female
                    
		</div>
				
            
            
            
               
	     Problem
              
				
            	visit
            	<?php
                     $query = "SELECT * from P_table";
		     $result = mysqli_query($con, $query);
		?>
		
     				Please select
		<?php
                    while ($row= mysqli_fetch_assoc($result)) {
		    $id = $row['u_id'];
		    $vist_to = $row['visit_to'];
		?>
	 <?php echo $vist_to;?>
	<?php } ?>
           	  
            		
            	
             	 Doctor:
                    Doctor Name
                      Dr.Henry
                      Dr.Menro
                    
                  <div class="input-group">
                    Fees:
                    <div class="input-group">
                    
		    
		   </div>
		</div>
              
	            
              <!-- Date and time range -->

              <div class="form-group">
                Date and time

                <div class="input-group">
                  <div class="input-group-addon">
                    
                  </div>
                  
                </div>
                <!-- /.input group -->
              </div>
              <!-- /.form group -->
                
            
            
            <div class="clear"></div>
                            
               
              <div class="clear"></div>         
	
	
<!--End of wrapper-->
?php
	if(isset($_POST['submit']))
	{
		$name = $_POST['name'];
		$f_name = $_POST['f_name'];
		$phone_no = $_POST['phone_no'];
		$country = $_POST['country'];
		$city = $_POST['city'];
		$add = $_POST['full_add'];
		$age = $_POST['age'];
		$age_sc = $_POST['age_sc'];
		$gender = $_POST['gender'];
		$prob = $_POST['problem'];
		$visit = $_POST['visit'];
		$doct = $_POST['doc'];
		$fee = $_POST['fee'];
		$crncy = $_POST['crncy'];
		$a_date = $_POST['date'];
		
		
		$insert = "insert into new_table (name,f_name,contact,country,city,address,age,age_sc,gender,problem,visit,doc,fee,currency,date,c_date) values('$name','$f_name','$phone_no','$country','$city','$add','$age','$age_sc','$gender','$prob','$doct','$fee','$crncy','$a_date',NOW())";
		
		$qur = mysqli_query($con, $insert);
		
		if($qur)
		{
		   echo"alert('Added Successfully')";
		   echo"window.open('index.php','_self')";
		   }
		else{
			echo "<h2>Data not inserted </h2>";
		}
		
	}
		?>

Posted 15-Jan-18 20:01pm

Member 13624902

Updated 15-Jan-18 23:13pm

Richard MacCutchan

v3

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Comments

Richard MacCutchan 16-Jan-18 4:06am

Try adding some debug statements to see what is happening.

1 solution

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Patrice T · Answer 1 · 2018-01-15T23:13:00

PHP

$insert = "insert into new_table (name,f_name,contact,country,city,address,age,age_sc,gender,problem,visit,doc,fee,currency,date,c_date) values('$name','$f_name','$phone_no','$country','$city','$add','$age','$age_sc','$gender','$prob','$doct','$fee','$crncy','$a_date',NOW())";

Never build an SQL query by concatenating strings. Sooner or later, you will do it with user inputs, and this opens door to a vulnerability named "SQL injection", it is dangerous for your database and error prone.
A single quote in a name and your program crash. If a user input a name like "Brian O'Conner" can crash your app, it is an SQL injection vulnerability, and the crash is the least of the problems, a malicious user input and it is promoted to SQL commands with all credentials.
SQL injection - Wikipedia[^]
SQL Injection[^]
SQL Injection Attacks by Example[^]
PHP: SQL Injection - Manual[^]
SQL Injection Prevention Cheat Sheet - OWASP[^]