PHP - Addslashes() Function .. Cannot Implement In My Mysql Insert - Why?

Hi guys

When I insert a data into MySQL and thru addslashes() it is adding not one but 3 slashes in mysql.

By the way here are the codes,
Code: [Select]
<?php
//$conn = new mysqli('localhost', 'root', '', 'my_db');
$conn = new mysqli('localhost', 'coder9_work', '******', 'coder9_portfolio');
$query = "INSERT into portfolio(category, title, description, version, started, finished) VALUES (?, ?, ?, ?, ?, ?)";

$select = $_POST['select'];
$title = addslashes($_POST['title']);
$description = $_POST['description'];
$version = $_POST['version'];
$started = $_POST['started'];
$finished = $_POST['finished'];


$stmt = $conn->stmt_init();
if($stmt->prepare($query)) {
$stmt->bind_param('ssssss', $select, $title, $description, $version, $started, $finished);
$stmt->execute();
}

if($stmt) {
echo "Thank you!";
} else {
echo "There was a problem. Please  try again later.";
}
?>

How do I fix this problem so that It will add only one slashes?

Thanks in advanced.

Hello, I'm having a bit of a problem here, all help to this issues would be much appreciated

I am trying to use text boxes to insert numbers into the database based on what is inputed.
If I have a string, like this for example:
$variable = 09385493;

And I want to insert it into the database like this:
mysql_query("INSERT INTO integers(number)
VALUES ('$variable')");

When checking the integers table in my database, looking at the number field, the $variable that was inserted is outputted as
9385493

Notice the number zero was taken out of the front of the number. If the number is double 0's (009385493), both of those zero's would disappear, too.

Thanks

I have a class built for an INSERT query but it is passing two sets of records into the database rather than one.

Code: [Select]
class DatabaseInsert {

function DatabaseConnectionRequire() {
include("../scrips/php/database.connection.class.php");
include("../scrips/php/database.settings.php");
include("../scrips/php/database.connection.class.invoke.php");

}

function ArticleInsert($values,$fields,$table) {

$values_imploded = implode("   ",$values);
$fields_imploded = implode("   ",$fields);
$i = "INSERT INTO $table ($fields_imploded) VALUES ($values_imploded)";
mysql_query($i) or die(mysql_error());

if (!mysql_query($i)) {

    echo "Sorry, something whent wrong there...";
}

else {
echo "<strong><p style='color:green;'>Content added sucessfully!!!</p></strong>";
}

}
}

HI All,
I would like create query when I add some one is look what the last number in field and add one ?

example :
you can see the img in attach
IDT is primary key + auto increment
Customer_id is number field
customer _name is name

how can create this login

IDT    Customer_ID    Customer_name
112           5                    bbbb
113           6                    ccccc
114           7                     eeeee

Also I need the inster function and is look the last field in customer_ID and Increase one ( +1)


Any help please/

THanks

HI,i am using java script to create a add row function in the php .but when the first row data can insert into database ,the 2nd row data cannot insert into database ,can help me to check my coding? thanks a lot


Code: [Select]
<SCRIPT language="javascript">
        function addRow(tableID) {
 
            var table = document.getElementById(tableID);
 
            var rowCount = table.rows.length;
            var row = table.insertRow(rowCount);
 
            var cell1 = row.insertCell(0);
            var element1 = document.createElement("input");
            element1.type = "checkbox";
            cell1.appendChild(element1);
 
         
            var cell2 = row.insertCell(1);
            var element2 = document.createElement("input");
            element2.type = "text";
            cell2.appendChild(element2);

var cell3 = row.insertCell(2);
            var element3 = document.createElement("input");
            element3.type = "text";
            cell3.appendChild(element3);

var cell4 = row.insertCell(3);
            var element4 = document.createElement("input");
            element4.type = "text";
            cell4.appendChild(element4);

var cell5 = row.insertCell(4);
            var element5 = document.createElement("input");
            element5.type = "text";
            cell5.appendChild(element5);

 
        }
 
        function deleteRow(tableID) {
            try {
            var table = document.getElementById(tableID);
            var rowCount = table.rows.length;
 
            for(var i=0; i<rowCount; i++) {
                var row = table.rows[i];
                var chkbox = row.cells[0].childNodes[0];
                if(null != chkbox && true == chkbox.checked) {
                    table.deleteRow(i);
                    rowCount--;
                    i--;
                }
 
            }
            }
catch(e) {
                alert(e);
            }
        }
 
    </SCRIPT>


THE PHP CODE
Code: [Select]
<?php 
require_once ('../../../Connections/admin_db.php'); 

mysql_select_db("admin_db");

if ((isset($_POST["Submit"])) && ($_POST["Submit"] == "Submit")) {
$i=0;
foreach($_POST['abc'] as $value )
{
   
                 $abc           = $_POST['abc'][$i];
 $level         = $_POST['level'][$i];
     $level_desc    = $_POST['level_desc'][$i];
      $pc_desc       = $_POST['pc_desc'][$i];
 
 
//Insert Data into Instructor Profile Info
$q = "INSERT INTO plo_pc(p_name,plo_id,plo_criteria,plo_level,level_dec,plo_desc)
VALUES ('$list','$plo_id','$abc','$level','$level_desc','$pc_desc') " ; 
 
mysql_query($q) or die(mysql_error()) ;
$i=$i+1;
}
}
?>

Hello all, so I created an insert function and it seems no matter what I try that it won't add values using the query function inside a table from the respective variables, I would like to know why is this happening? Here is the code can you tell me why it doesn't insert anything in the database? It shows no errors when it runs but then again when I check the tables they're empty!

function insert(){
				$user = $_POST['user'];
				$pass = md5($_POST['pass']);
				$priv = "User";
				$mail = $_POST['mail'];
				$avatar = $_FILES['avatar']['name'];
				$date="now()";
				$submit = $_POST['submit'];
			
			
				$query = "INSERT INTO user(user,pass,priv,mail,avatar,date) VALUES(`$user`,`$pass`,`$priv`,`$mail`,`$avatar`,`$date`);";
				if($submit){
					$res = mysqli_query($con,$query) or die(mysqli_error($con));
				}
				
				
			}

Hi guys

I have a registration form working fine, my database is as below:

userid
username
email
password
repeatpassword


I have added another column which is "name", users can update their profile once they have logged in so I have created updateprofile.php and when I login-->go to update profile and insert my name nothing adds to mysql name column

this is my code below:

<?php

include ("global.php");


//username session
$_SESSION['username']=='$username';
$username=$_SESSION['username'];

//welcome messaage
echo "Welcome, " .$_SESSION['username']."!<p>";

  if ($_POST['register'])
    {
     //get form data
     $name = addslashes(strip_tags($_POST['name']));

$update = mysql_query("INSERT INTO users (name) VALUES ('$_POST[name]') WHERE username='$username'");
   
}

?>

    <form action='updateprofile.php' method='POST'>
    Company Name:<br />
    <input type='text' name='name'><p />
 
    <input type='submit' name='register' value='Register'>
    </form>


can you please tell me where in this code is wrong? Im new in php so please excuse me if I have silly mistakes.

thanks in advance
 

I have this code: <?php
$con = mysql_connect("localhost","hhh","hhh");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("hhh", $con);

// --------------------
// Avatar insert check
// --------------------

session_start();

$name = $_POST[name];
$group = $_POST[group];
$age = $_POST[age];
$usernameid = $_SESSION[id];

$result = mysql_query("SELECT * FROM avatars WHERE name='$_POST[name]'");

$num = mysql_numrows($result);

if ($num == 0) {
mysql_query("INSERT INTO avatars (id, usernameid, name, group, age, xp)
VALUES ('', '$usernameid', '$name', '$group', '$age', '0')");

header( 'Location: me/' ) ;
}
else echo 'Sorry, please pick a new name';
?> And it does everything but put the data into the datebase. If I add a session befor and after '$request' they both run, but the sql doesn't. No error returns, if just redirects to the other page. Any help?

well this is truely embarrising...i have a insert statement which works within phpmyadmin but when using mysqli_query it returns a error.

INSERT INTO users (username, timestamp) VALUES ('test', UTC_TIMESTAMP())

  Unknown column 'timestamp' in 'field list'   i've been playing about with this for a few hours now ...tried changing the column name (timestamp), adding ` around column names as well as table name.   the column exists which is the strangest part, and ive even checked there is no space after the column name in the db.   whats going on please?  

I don't understand where the empty value is. I've substituted the variables for text and still have the same problem.

Code:
Code: [Select]
$sql = "INSERT INTO courses (course#, name, subject, semester, ap)VALUES('$courseNum', '$courseName', '$subject', '$semester', '$ap')";
Error:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

I need help badly!

What I want to do is insert into database the value from the selected radio group buttons.. All of them. There are 10 radio groups total (they can be less, but not more).

Thanks!
Code: [Select]
<?php
   require_once('Connections/strana.php'); 
 mysql_select_db($database_strana, $strana);
 
 ?>
<link href="css/styles.css" rel="stylesheet" type="text/css" />

<table width="100%" height="100%" style="margin-left:auto;margin-right:auto;" border="0">
<tr>
<td align="center">
<form action="" method="post" enctype="multipart/form-data" name="form1">
<table>
<?php
$tema = mysql_query("SELECT * from prasanja where tip=2")or die(mysql_error());

function odgovor1($string)
{
$string1 = explode("/", $string);
echo $string1[0];
}
function odgovor2($string)
{
$string1 = explode("/", $string);
echo $string1[1];
}


while ($row=mysql_fetch_array($tema))
{
$id=$row['prasanje_id'];
$prasanje=$row['prasanje_tekst'];
$tekst=$row['odgovor'];
?>
    <tr>
    <td>
    </td>
    </tr>
    <tr>
    <td class="formaP">
 <?php echo $prasanje?>
   </td>
   </tr>
   <tr>
   <td class="formaO">
      <p>
        <label>
          <input type="radio" name="Group<?php echo $id?>" value="<?php odgovor1($tekst) ?>"  />
          <?php odgovor1($tekst) ?></label>
        <br />
        <label>
          <input type="radio" name="Group<?php echo $id?>" value="<?php odgovor2($tekst) ?>" />
          <?php odgovor2($tekst) ?></label>
        <br />
      </p></td>
</tr>
<tr>
<td>
<br />

</td>
</tr>
<?php 




}
?>
</table>
<input align="left"type="submit" name="submit"
value="Внеси" >
</form>
</td>
</tr>
</table>
prasanje = question
tekst/odgovor = answer

The answer table:
id - primary
question_id - the questions ID whose answer is selected in the radio group
user_id - cookie takes care of this
answer - the value from radio group
date - automatic

Is there any way to tell my php ajax file to run the update query if the data already exist and if not, then create the row in the database?  I have both the update and the insert functions created, but was just wondering if I could tell php which one to use without passing through a parameter.

This topic has been moved to JavaScript Help.

http://www.phpfreaks.com/forums/index.php?topic=308768.0

My hosting service has magic_quotes_gpc = On.  I was working on my home test server and the following script worked perfectly.  Turns out I had magic_quotes_gpc = Off .. I set magic_quotes_gpc = On and restarted.  Now the script isn't working.  See code and output below.  I know something isn't being escaped properly, but I have no clue how/what.  Even if I copy and paste the $insert output directly to phpmyadmin, it returns the same error.

Code: [Select]
//HTML Vars
$firstName = $_POST['firstName'];
$lastName = $_POST['lastName'];
$email = $_POST['email'];
$desc = $_POST['desc'];    //This is a textarea with long description.
$year = $_POST['date'];

//MySQL - no connection issues
$link = mysql_connect('localhost', '__uesr__', '__passwd__*');
$db = mysql_select_db('__DB__', $link);

$insert = "INSERT INTO images (firstName, lastName, email, descript, dateYear) VALUES ('$firstName' , '$lastName' , '$email' , '$desc' ,  '$year' ";

$query = mysql_query($insert);
if (!$query) {
die ('Can\'t query ' . mysql_error());
}

echo $insert;

::OUTPUTS::

Can't query You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

INSERT INTO images (firstName, lastName, email, descript, dateYear) VALUES ('this' , 'is' , 'the@email.com' , 'and. the. description won\'t work.' , '3456'



Obviously I am a novice.  I have tried using mysql_real_escape_string with and without stripslashes, but I'm not getting anywhere except more errors.  Any help would be greatly appreciated.  And I don't care about SQL injection AT ALL.  I just want the thing to work with proper escaping for the description if a user inputs special chars.