Question

Dumping Data into mySQL using PHP

Asked by: sam928

hello, i have a PHP script that creates a table in mySQL.  

QUESTION.  how do i dump data into the database with the table?

below is my script.

$result="CREATE TABLE ".$albumName." (id int(11) unsigned NOT NULL AUTO_INCREMENT,PRIMARY KEY ( `id` ),albumName varchar(100) default NULL,imageName varchar(100) default NULL,imageText varchar(100) default NULL)";

                                  
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Asked On
2009-11-07 at 10:33:32ID24880724
Tags

php

,

mySQL

Topics

PHP and Databases

,

PHP Scripting Language

Participating Experts
3
Points
500
Comments
10

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Answers

 

by: Ray_PaseurPosted on 2009-11-07 at 10:45:53ID: 25767503

Sometimes the question reveals some missing pieces of the knowledge foundation, and this is one of those times.  This question might as well be, "Here is my propeller.  How do I fly an airplane?"  Let's see if we can get you onto a firm footing with PHP.

A really good starting place.
http://us3.php.net/tut.php

After that, please continue your study here:
http://www.php.net/docs.php

Looking especially at this:
http://www.php.net/links.php

A really good book that will teach you the answer to this question and much more is available from SitePoint:
http://www.sitepoint.com/books/phpmysql4/

Best of luck with your project, ~Ray

 

by: angelIIIPosted on 2009-11-07 at 10:47:03ID: 25767509

where does the data come from?

 

by: rpkharePosted on 2009-11-07 at 11:36:16ID: 25767701

Well, as said above, your question 'n concept is not clear. Still, with what I understand, you are a bit confused with a Table and a Database. Since you have mentioned MySQL, let me clear few things below. I am not sure whether you are desiring this or not.

(1) A MySQL database consists of tables where in you store records. After creating structure of your table, you need to fire Insert query to insert records in the table, that is what you are calling dumping. But, you are not dumping table, but you are inserting records in the tables which themselves reside inside the database.

For this, refer this link: http://www.w3schools.com/PHP/php_mysql_insert.asp

(2) A MySQL dump is actually a MySQL database backup.

For this refer this link: http://dev.mysql.com/doc/refman/5.1/en/mysqldump.html


 

by: sam928Posted on 2009-11-08 at 08:35:51ID: 25770929

@Ray. thank you for that information.
---

@rpkhare..  sorry for the mix up.. i am looking for insert records.. not 'dump' them.

i have viewed the first link and tried to build the example in my script but was unsuccessful... Below is the entire PHP..

....starts on line 8  ...WHat am i doing wrong here?  (everything else works EXCEPT the inserting of the test record.

$albumName = $_POST['albumName'];
//$albumName = $_POST['albumName'];
 
 
$result="CREATE TABLE ".$albumName." (id int(11) NOT NULL AUTO_INCREMENT,
PRIMARY KEY(id),albumName varchar(100) default NULL,imageName varchar(100) default NULL,imageText varchar(100) default NULL)";  
mysql_query("INSERT INTO $albumname (albumName, imageName, imageText)
VALUES ('test', 'test2', 'test3')");
 
 
if(mysql_query($result))
{
   echo("Error:".mysql_error()." For query:".$result);
} else { 
if($result) echo "writing=Ok&";
   else echo "writing=Error";
   }
   
   
   //$albumName = "testdir";  
mkdir("http://www.previewsample.com/album/" . $albumName . "/"); 
if(mkdir($albumName)){
echo 'successfully created directory';
} else {
die('problem creating directory'); 
}
?>  
                                              
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by: rpkharePosted on 2009-11-08 at 08:55:06ID: 25770978

Where you are opening connection to MySQL? Also to debug, set:

echo $result;

just after the line $result to see if you are using concatenated values correctly.

 

by: sam928Posted on 2009-11-08 at 09:20:03ID: 25771051

i did not include my connection lines..
--
i inserted that and tested in browser.. it said this:

CREATE TABLE (id int(11) NOT NULL AUTO_INCREMENT, PRIMARY KEY(id),albumName varchar(100) default NULL,imageName varchar(100) default NULL,imageText varchar(100) default NULL)writing=Ok&problem creating directory

 

by: sam928Posted on 2009-11-08 at 10:09:42ID: 25771235

ok, let me start this over.. it is quite simple.. all i want to do is insert some data (record) into the created database.

THE ENTIRE SCRIPT BELOW WORKS FINE.

it does the following:

1) creates a table with the variabe
2) creates a directory on the server with the variable name.

-----------------------
All i want to do is insert a record into the newly created table.
-----------------

DOES ANYBODY KNOW HOW TO DO THIS WITHOUT BEING CONFUSING..?   just a simple simple example will work. ;)

I think alot of people get carried away on this site trying to show off their skills by suggesting over complex solutions.. it defeats the purpose of the person asking the question trying to learn.

<?php
$connect = mysql_connect("dxxxxx", "dxxxxxxx", "gxxxxxxxxx");
   mysql_select_db ("xxxxxxxxxx", $connect); 
 
 
$albumName = $_POST['albumName']; 
 
 
$result="CREATE TABLE ".$albumName." (id int(11) NOT NULL AUTO_INCREMENT,
PRIMARY KEY(id),albumName varchar(100) default NULL,imageName varchar(100) default NULL,imageText varchar(100) default NULL)"; 

$sql = "INSERT INTO users (name,city,web,age) VALUES ('Tom','Vegas','www.tom.com',5)";
$result = mysql_query($sql); 
mysql_query("INSERT INTO $albumName (albumName, imageName, imageText)
VALUES ('Peter', 'Griffin', '35')"); 
if($result) echo "writing=Ok&"; 
else echo "writing=Error";  
if(mysql_query($result)) 

mkdir("http://www.previewsample.com/album/" . $albumName . "/",0777); 
if(mkdir($albumName)){
echo 'successfully created directory';
} else {
die('problem creating directory'); 
}
?>  

                                              
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by: sam928Posted on 2009-11-08 at 10:13:21ID: 25771242

please disregard the above PHP..  

below is the current working script.. Where would I add the line to INSERT A RECORD?

<?php
$connect = mysql_connect("dcccccc", "cccccccccc", "ccccccc");
   mysql_select_db ("cccccccccc", $connect); 
 
 
$albumName = $_POST['albumName']; 
 
 
$result="CREATE TABLE ".$albumName." (id int(11) NOT NULL AUTO_INCREMENT,
PRIMARY KEY(id),albumName varchar(100) default NULL,imageName varchar(100) default NULL,imageText varchar(100) default NULL)"; 
  
if($result) echo "writing=Ok&"; 
else echo "writing=Error";  
if(mysql_query($result))  
mkdir("http://www.previewsample.com/album/" . $albumName . "/",0777); 
if(mkdir($albumName)){
echo 'successfully created directory';
} else {
die('problem creating directory'); 
}
?>       

                                              
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by: angelIIIPosted on 2009-11-08 at 12:04:24ID: 25771652

>Where would I add the line to INSERT A RECORD?
with what data?

as in your other question, it looks like you want to create 1 table per album, instead of have 1 table (created once and for all), and insert 1 record/row into that table.

let's say you create the table with the name ALBUMS, with the same structure you showed.

<?php
$connect = mysql_connect("dcccccc", "cccccccccc", "ccccccc");
   mysql_select_db ("cccccccccc", $connect); 
 
 
$albumName = mysql_real_escape_string($_POST['albumName']); 
$imageName = mysql_real_escape_string($_POST['imageName']); 
$imageText = mysql_real_escape_string($_POST['imageText']); 
 
 
$result="INSERT INTO ALBUS(albumName, imageName,imageText ) VALUES ('$albumName', '$imageName', '$imageText') "; 
  
if($result) echo "writing=Ok&"; 
else echo "writing=Error";  
if(mysql_query($result))  
mkdir("http://www.previewsample.com/album/" . $albumName . "/",0777); 
if(mkdir($albumName)){
echo 'successfully created directory';
} else {
die('problem creating directory'); 
}
?>       

                                              
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by: sam928Posted on 2009-11-09 at 14:48:23ID: 31651430

Thank you all for the insight..

I have found a complete solution for this on another question.

Thank you.

20120131-EE-VQP-002

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