Using Session Variables To Insert Data To MYSQL Database?

May 23, 2009

I am having problems getting PHP to insert data into a MYSQL data base. I am using the following variations of syntax.

mysql_query("INSERT INTO table member VALUES ('$var1', '$var2', '$var3')")
or die("couldn't add new user");

and also with session variables as well

mysql_query("INSERT INTO table member VALUES ('$_SESSION[$var1]', '$_SESSION[$var2], '$_SESSION[$var3]')")or die("couldn't add new user");

i have put all these through a print statement to make sure the variables ae set and they print in the same block of code so i have no idea what i am doing wrong.?

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Insert Query Doesn't Insert Data To Mysql Database Table?

i have a recipe table and ingredient table the primary key of both tables are auto increament and primary key of recipe is foreign key in ingredient. i post data from html to php.Note that my ingredient textboxes are generated dynamically and successfully post the data to php script. posted data is correct when i insert this data to table my query working fine but data is not added to mysql table. my code and output is

$sql = "insert into recipe (rec_id, Name, Overview,category, Time, Image) values ('', '$name','$overview','$category','$time','$TARGET_PATH')";
$result = mysql_query($sql) or die ("Could not insert data into DB: " . mysql_error());
$rec_id = mysql_insert_id();


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I need to INSERT data into my MySQL database depending on whether a field in another table in the database is TRUE or FALSE (bool). The PHP code I have so far goes something like this:

//Insert data into database
mysql_query("INSERT INTO answers (name, answer, aQID, date)
VALUES ('$name', '$answer', '$curdate')");

I am wondering where I could put the WHERE clause. It needs to be something like this

WHERE approvedQuestions status = TRUE

I am hoping I'm explaining myself clearly. Since I am not displaying data I do not believe I need to be using the UNION but I could be wrong.

EDIT: Each question has a unique ID associated with it. The unqiue ID is also present in the answer table for association purposes. Only thing is all the relationships are being done PHP wise and not MySQL wise. What would be the best way to assign the unique questionID to the answer table?

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I am working on a memebship section for a website and I am trying to set session variables from records in a MySql database when the user logs in. I think I am missing something because try as I might the just don't work!

The sever I use had Glabal variables turned on, dont know if thats an issue in this case.

Heres the code on the check user page:

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MySQL / Insert The Data Into The Database?

I basically made a registration form and need to insert the data into the database.

I am not sure how i use the MySQL functions to do this,

I created a db connection file and included it on registration page but not sure if i first havew to query database then insert data etc, i am a little confused.

Before any data can be inserted it has to check if the email address and username already exist. Could someone plese give me the info on what order i do it?

here is what needs being added to database.

username, first_name, last_name, email, password

before it inserts the users data it needs to check to make sure the username and email does or does nto exist.

Here is my db connection file code in case it helps, this is included using the php include in the registration page. All i need help in how i got about checkning if username and email exists or not and if all ok put info in db or give an error to say it already exists.


// Database Connection Details
$host = "localhost"; // Usually Local Host
$user = "none"; // Database Username
$pass = "none"; // Database Password
$database = "usernone"; // Database Name
mysql_connect ($host, $user, $pass, $database);

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I'm doing a tool for a game, and I need to insert some data to my MySQL database.I use this code, but it dont work

include 'config.php';
include 'checklogin.php';
include 'main.php';


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I'm having trouble adding data to the database (test) that i created. I just can't see where I'm going wrong. need some other peoples opinions.I have tried taking the id out of the SQL statement completely as it is an auto increment, but that doesn't seem to work either.

$fname = $_POST['fname'];
$sname = $_POST['sname'];

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Retrieve MySql Data And Store In Session Variables?

I feel like I'm blowing this forum up today with random topics but I really don't want to throw it all in one post after the questions have been answered.That being said, I think I'm down to my last question of the day...My session stuff is finally working....yay!....but I have one last question...I know I can assign session['name'] with = and a direct quote but I'm having problems getting the query data to populate this raises two questions with me...1. in my_sql, the array should start at 0, right?2. When I'm assigning the values to my session variables, how limited are my options? Are these variables standard or can I make them up as I go? meaning if I were to use $_SESSION['name'] or $_SESSION['username'] to assign the users name to, from the db, would i be able to get the same result?

@mysql_select_db($database) or die( "Unable to select database");
$query="SELECT * FROM vis_master_links WHERE linktype=1 ORDER BY linktype, linkorder";

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Use A Form To Insert Data Into A MySQL Database

When you use a form to insert data into a MySQL database, how can you
set it up so that the fields are not sent via the address bar.

For example:

<form action="page2.php">
<input type="text" value="whatever">
<input type="submit" value="Send">

// Inserts variable $whatever into database.

The problem that I have is that I don't want the address bar to appear
as http://web/page2.php?whatever=cheese but as http://web/page2.php

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I have run createTable.php and everything is created. Now I want to run insertData.php. I want the user to enter stuff in the form & everything must be fill in before they click the submit button. It will call submitData() & all the information was enter will insert into the database. PHP Code:

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I have a textarea that is part of a HTML form and I am trying to insert the data from this textarea "comments" into a MySQL database.There are other inputs at seem to be working but for some reason the comments one doesn't. Here is the code for the textarea in the form:

<textarea type='text' name="comments">

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i want to insert into mysql database from xls file using php.

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i have copied this example php code to insert data and a picture into a mysql database. however i get this error.Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource [URL]....

Code: [Select]<?php
$db_host = 'xxx'; // don't forget to change
$db_user = 'xxx';

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I am trying to display data from mysql. Each row from the mysql database shows up under rows. I want the rows of data from mysql to appear in colums of 4.

$sql = 'SELECT * FROM dbtable ORDER BY id';
$result = $db->query($sql);
$output[] = '<ul>';
while ($row = $result->fetch()) {
$output[] = '<table>';
$output[] = '<tr>';
$output[] = '<td class="style10"><strong>'.$row['item'].'</strong></td>';

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When I try to insert data into my MySQL database I get an error that it cannot connect to my DB. I have it setup so why does it not work? I have WampServer installed.

<form method="post" action="update.php">
First Name:<br/>
<input type="text" name="FName" size="30" /><br/>
Last Name:<br/>
<input type="text" name="LName" size="30" /><br/>
<input type="text" name="phone" size="12" /><br/>

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I want to collect the data from an XML feed to insert into a MySQL database, but I can't seem to get the Node value of an attribute to insert in to a database, as it is an object and not a string although the variable will echo and print. I can not seem to convert this to a string.

$years = $findit->getElementsByTagName("vehicle")->item(0);
$year = $years->getAttribute('year');

This produces... Catchable fatal error: Object of class DOMAttr could not be converted to string. But the element by tag name works without any problem.

$manufacturers = $findit->getElementsByTagName( "manufacturer" );
$manufacturer = $manufacturers->item(0)->nodeValue;

why the element object and attribute object work differently.

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I am using a Mysql database and I'm trying to retrieve form data and insert it into a MySQL database, which I can do for all my standard text boxes, However, I am having issues with my textarea.

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I have created a form to insert data into a MySQL database and I now keep getting an error message. ere is my injection string:

$user = mysql_query("insert into users(user_name,password,name,city,address,email,country_symbol) values('".$_REQUEST['user_name']."','".$_REQUEST['password']."','".$_REQUEST['name']."','".$_REQUEST['town']."','".$_REQUEST['address']."','".$_REQUEST['email']."','".$_REQUEST['country']."')")or die(mysql_error());

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PHP Code:
* File : session.lib..php
* Project : Sessions
* Author : AD
* Description : Sessions stored into the database

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From a user form: I am trying to insert the following data: 1) First Name 2) Last Name 3) Major 4) Graduation Year

I am able to connect to the database, and select the database I need--but I am unable to insert the data from the form. I am able to create records, but the data is not being saved to the database. Basically, right now I'm creating blank forms.

The variable $uInput holds the user data. I tried passing $uInput into the function doAction(), but I believe that is where the problem is. I'm trying to figure out how to pass the user data into the function doAction().

//Call function mainline
// Declare the function mainline
function mainline() {


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class Order extends AppModel {
var $name="Order";
var $belongsTo = array( [code]....

and I have created relation in database that foreign key studentId of order table references Id column of Student table. So now how to save both associated models?

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$query = "SELECT * FROM users WHERE name = '$username'";
$result = mysql_query($query, $connection) or die('error making query');

while ($row = mysql_fetch_array($result, MYSQL_BOTH)) {
$organization = $row["organization"];
$contactperson = $row["contactperson"];}

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