Solved

why i cant insert into a field in a table in mysql things like "."  "?"   "," ?????

Posted on 2008-06-10
8
235 Views
Last Modified: 2012-05-05
i have a field in a table and i want to insert in it this symbols
?
.
,
'
but its not working
why is that?
0
Comment
Question by:mgtm3
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
8 Comments
 
LVL 5

Accepted Solution

by:
vibrazy earned 250 total points
ID: 21749453
Try using mysql_real_escape_string($fieldname);
0
 

Author Comment

by:mgtm3
ID: 21749461
look att the code and tell me what to do?
<?php
 
include 'cpoll.php';
 
 
$subject = $_POST['subject'];
$op1 = $_POST['op1'];
$op2 = $_POST['op2'];
$op3 = $_POST['op3'];
$op4 = $_POST['op4'];
$op5 = $_POST['op5'];
$op6 = $_POST['op6'];
$op7 = $_POST['op7'];
$op8 = $_POST['op8'];
$op9 = $_POST['op9'];
$op10 = $_POST['op10'];
$op11 = $_POST['op11'];
$op12 = $_POST['op12'];
$gen=$_POST['gen'];
 
 
$result = mysqli_query($link, "SELECT count(*) FROM $gen");
$row = mysqli_fetch_array($result, MYSQLI_NUM);
$row[0]++;
$pollname="poll".$gen.$row[0];
echo $pollname; 
 
 
 
echo "subject - ".$subject."<br />op1 - ".$op1."<br />op2 - ".$op2."<br />op3 - ".$op3."<br />op4 - ".$op4."<br />op5 - ".$op5."<br />gen - ".$gen;
 
$sql = "CREATE TABLE $pollname (
`id` int( 255 ) NOT NULL AUTO_INCREMENT,
  `subject` VARCHAR( 700 )NOT NULL,
        `op1` VARCHAR( 700 )NOT NULL,
        `op2` VARCHAR( 700 ) NOT NULL,
        `op3` VARCHAR( 700 ) NOT NULL,
        `op4` VARCHAR( 700 ) NOT NULL,
        `op5` VARCHAR( 700 ) NOT NULL,
		 `op6` VARCHAR( 700 )NOT NULL,
        `op7` VARCHAR( 700 ) NOT NULL,
        `op8` VARCHAR( 700 ) NOT NULL,
        `op9` VARCHAR( 700 ) NOT NULL,
        `op10` VARCHAR( 700 ) NOT NULL,
		`op11` VARCHAR( 700 ) NOT NULL,
        `op12` VARCHAR( 700 ) NOT NULL,
            `date` VARCHAR( 700 ) NOT NULL,
            `user` VARCHAR( 700 ) NOT NULL,
			            `comment` text NOT NULL,
 
            `gen` VARCHAR( 700 ) NOT NULL,
            
        PRIMARY  KEY ( `id` )
       )";
 
 
 
 
 
mysqli_query($link,$sql);
 
mysqli_close($link);
unset($link);
include 'cpoll.php';
 
$f = "INSERT INTO $pollname (op1,op2,op3,op4,op5,subject,gen) VALUES ('$op1','$op2','$op3','$op4','$op5','$subject','$gen')"; 

Open in new window

0
 
LVL 5

Expert Comment

by:vibrazy
ID: 21749493
Try this, if that doesnt work, tell me what error you are getting.

Regards,
Vibrazy
$f = "INSERT INTO $pollname (op1,op2,op3,op4,op5,subject,gen) VALUES ('".$op1."','".$op2."','".$op3."','".$op4."','".$op5."','".$subject."','".$gen."')"; 
mysqli_query($link,$f);

Open in new window

0
Comprehensive Backup Solutions for Microsoft

Acronis protects the complete Microsoft technology stack: Windows Server, Windows PC, laptop and Surface data; Microsoft business applications; Microsoft Hyper-V; Azure VMs; Microsoft Windows Server 2016; Microsoft Exchange 2016 and SQL Server 2016.

 

Author Comment

by:mgtm3
ID: 21749508
can u write me the function to get the error ?
0
 

Author Comment

by:mgtm3
ID: 21749540
o.k this what i got

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 's Day','Valentine's Day','Favorite holiday?','other')' at line 1
0
 
LVL 28

Assisted Solution

by:gamebits
gamebits earned 250 total points
ID: 21749671
$subject = $_POST['subject'];
$op1 = mysql_real_escape_string ($_POST['op1']);
$op2 = mysql_real_escape_string ($_POST['op2']);
$op3 = mysql_real_escape_string ($_POST['op3']);
$op4 = mysql_real_escape_string ($_POST['op4']);
$op5 = mysql_real_escape_string ($_POST['op5']);
$op6 = mysql_real_escape_string ($_POST['op6']);
$op7 = mysql_real_escape_string ($_POST['op7']);
$op8 = mysql_real_escape_string ($_POST['op8']);
$op9 = mysql_real_escape_string ($_POST['op9']);
$op10 = mysql_real_escape_string ($_POST['op10']);
$op11 = mysql_real_escape_string ($_POST['op11']);
$op12 = mysql_real_escape_string ($_POST['op12']);
$gen = mysql_real_escape_string ($_POST['gen']);


Now you can use it in a query
0
 

Author Comment

by:mgtm3
ID: 21749716
thanksssssssssssssssss
0
 
LVL 4

Expert Comment

by:afzz
ID: 21749726
You can also do the following instead of individually changing the values
foreach($_POST as $key => $val){
${$key} = mysql_real_escape_string($val);
}

Open in new window

0

Featured Post

Microsoft Certification Exam 74-409

Veeam® is happy to provide the Microsoft community with a study guide prepared by MVP and MCT, Orin Thomas. This guide will take you through each of the exam objectives, helping you to prepare for and pass the examination.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Suggested Solutions

Title # Comments Views Activity
MySQL recovery 7 28
How Close unsubmited attempts 10 43
Echo'd values in dropdowns 6 28
PHP $_POST vs asp request 4 22
As a database administrator, you may need to audit your table(s) to determine whether the data types are optimal for your real-world data needs.  This Article is intended to be a resource for such a task. Preface The other day, I was involved …
Nothing in an HTTP request can be trusted, including HTTP headers and form data.  A form token is a tool that can be used to guard against request forgeries (CSRF).  This article shows an improved approach to form tokens, making it more difficult to…
The viewer will learn how to count occurrences of each item in an array.
The viewer will learn how to create and use a small PHP class to apply a watermark to an image. This video shows the viewer the setup for the PHP watermark as well as important coding language. Continue to Part 2 to learn the core code used in creat…

735 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question