Dynamic Input Error To Mysql Using PHP

Question

Here is the database

CREATE TABLE IF NOT EXISTS `student` (
  `id` int(11) NOT NULL,
`name` varchar(100) NOT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;

ALTER TABLE `student`
ADD PRIMARY KEY (`id`);
ALTER TABLE `student`
MODIFY `id` int(11) NOT NULL AUTO_INCREMENT;

Here is the HTML part: jsFiddle

And Here is the PHP part inserting data to mysql:

<?php
$connect = mysql_connect('localhost','root','');
if (!$connect) {
   echo mysql_error();
}
$db = mysql_select_db('demo');
if (!$db) {
  echo mysql_error();
} 
$all_names ="";
if(isset($_POST["mytext"])){   
   foreach($_POST["mytext"] as $key => $text_field){
      $all_names .= $text_field .", ";
 }
}

$result = "INSERT INTO student ( name ) VALUES( $all_names )";
$insert_row  = mysql_query($result);
if(!$insert_row){
   echo $result;
}

Getting Error inserting value to database:

INSERT INTO student ( name ) VALUES( dffd, )

why Question with jQuery in title is not tagged with jQuery tag? and has no jQuery code included? — Alex
– Alex, Commented Oct 14, 2015 at 18:52
change echo mysql_error(); to echo $sql; and show us results — Grzegorz Adam Kowalski
– Grzegorz Adam Kowalski, Commented Oct 14, 2015 at 18:56
@nikolas, of course it shows sql query. This is exactly what I need to see. — Grzegorz Adam Kowalski
– Grzegorz Adam Kowalski, Commented Oct 14, 2015 at 18:59
@nikolas, edit your question and paste this resulting query. — Grzegorz Adam Kowalski
– Grzegorz Adam Kowalski, Commented Oct 14, 2015 at 19:00

Alex · Accepted Answer · 2015-10-14 19:12:10Z

2

I don't see whole picture yet. What is the real value returned by $_POST["mytext"]?

But my guess you should at least change your loop to:

foreach($_POST["mytext"] as $key => $text_field){
      if ($all_names == '') {
          $all_names .=  "( '".mysql_real_escape_string($text_field) ."') ";
      } else {
          $all_names .=  ",( '".mysql_real_escape_string($text_field) ."') ";
      }
 }

and your query to:

 $sql = "INSERT INTO student ( name ) VALUES $all_names ";

edited Oct 14, 2015 at 19:12

answered Oct 14, 2015 at 19:02

Alex

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3 Comments

Russell Over a year ago

Same error: INSERT INTO student ( name ) VALUES ,( 'dffd')

Russell Over a year ago

Comma is creating problem.

Alex Over a year ago

my bad typo in if ($all_names == '') {

Grzegorz Adam Kowalski · Accepted Answer · 2015-10-14 19:14:41Z

1

Change

   foreach($_POST["mytext"] as $key => $text_field){
      $all_names .= $text_field .", ";
   }

to

$all_names = "'" . implode("'), ('", $_POST["mytext"]) . "'";

answered Oct 14, 2015 at 19:14

Grzegorz Adam Kowalski

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6 Comments

Grzegorz Adam Kowalski Over a year ago

If you'd like to sanitize input then you could use array_map and mysql_real_escape_string.

Russell Over a year ago

is it like $all_names = "'" . implode("'), ('", mysql_real_escape_string($_POST["mytext"])) . "'";

Grzegorz Adam Kowalski Over a year ago

No, something like $all_names = "'" . implode("'), ('", array_map('mysql_real_escape_string', $_POST["mytext"])) . "'";

Russell Over a year ago

Many Many thanks again. Another question: if I have another single value with it. How to save that field.

Grzegorz Adam Kowalski Over a year ago

I do not understand. You want to INSERT not one, but two values into the table? Then probably it would be easier for you to modify @Alex code, but you could also continue using implode with earlier call to array_walk..

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