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PHP - Php Insert Info Mysql Table Query

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Php Insert Info Mysql Table Query	View Content

How can i use single quote for values?

$qry='insert into tablename values('a','b');';

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Mysql Database Insert Validation, And Displaying Info

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Hi,

I have a website where users can log on and edit their profile pic, name, biography etc.
I was wondering about the correct way to:-
Add data to the database through forms (Register.php) Display the data on a page Using mysql escape sting, however, the way I am currently using will display a '\' before any ' symbol. So it's >> it\'s ...

Here is a snippet of the code I am using...

Code: [Select]
//insert data
$about1 = mysql_real_escape_string($_POST['about']);

//get
$query = mysql_query("SELECT * FROM `staff` WHERE username='$username'");
$row = mysql_fetch_array($query);
$about = $row['about'];
echo $about;

Mysql Query Doesn't Work. [insert Query]

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If you also have any feedback on my code, please do tell me. I wish to improve my coding base.

Basically when you fill out the register form, it will check for data, then execute the insert query. But for some reason, the query will NOT insert into the database. In the following code below, I left out the field ID. Doesn't work with it anyways, and I'm not sure it makes a difference.

Code:

Code: [Select]
mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)");
Full code:

Code: [Select]
<?php
include_once("includes/config.php");

?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title><? $title; ?></title>
<meta http-equiv="Content-Language" content="English" />
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<link rel="stylesheet" type="text/css" href="style.css" media="screen" />
</head>
<body>

<div id="wrap">

<div id="header">
<h1><? $title; ?></h1>
<h2><? $description; ?></h2>
</div>

<? include_once("includes/navigation.php"); ?>

<div id="content">
<div id="right">
<h2>Create</h2>
<div id="artlicles">
<?php

if(!$_SESSION['user'])
{

        $username = mysql_real_escape_string($_POST['username']);
        $password = mysql_real_escape_string($_POST['password']);
        $name = mysql_real_escape_string($_POST['name']);
        $server_type = mysql_real_escape_string($_POST['type']);
        $description = mysql_real_escape_string($_POST['description']);

        if(!$username || !$password || !$server_type || !$description || !$name)
        {

        echo "Note: Descriptions allow HTML. Any abuse of this will result in an IP and account ban. No warnings!<br/>All forms are required to be filled out.<br><form action='create.php' method='POST'><table><tr><td>Username</td><td><input type='text' name='username'></td></tr><tr><td>Password</td><td><input type='password' name='password'></td></tr>";
        echo "<tr><td>Sever Name</td><td><input type='text' name='name' maxlength='35'></td></tr><tr><td>Type of Server</td><td><select name='type'>

        <option value='Any'>Any</option>
        <option value='PvP'>PvP</option>
        <option value='Creative'>Creative</option>
        <option value='Survival'>Survival</option>
        <option value='Roleplay'>RolePlay</option>

        </select></td></tr>

        <tr><td>Description</td><td><textarea maxlength='1500' rows='18' cols='40' name='description'></textarea></td></tr>";
        echo "<tr><td>Submit</td><td><input type='submit'></td></tr></table></form>";
        }
        elseif(strlen($password) < 8)
        {
        echo "Password needs to be higher than 8 characters!";
        }
        elseif(strlen($username) > 13)
        {
                echo "Username can't be greater than 13 characters!";
        }
        else
        {
                $check1 = mysql_query("SELECT username,name FROM servers WHERE username = '$username' OR name = '$name' LIMIT 1");

                if(mysql_num_rows($check1) < 0)
                {
                        echo "Sorry, there is already an account with this username and/or server name!";
                }
                else
                {
                        $ip = $_SERVER['REMOTE_ADDR'];

                        mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)");
                        echo "Server has been succesfully created!";
                }
        }

}
else
{
        echo "You are currently logged in!";
}

?>
</div>
</div>

<div style="clear: both;"> </div>
</div>

<div id="footer">
<a href="http://www.templatesold.com/" target="_blank">Website Templates</a> by <a href="http://www.free-css-templates.com/" target="_blank">Free CSS Templates</a> - Site Copyright MCTop
</div>
</div>

</body>
</html>

Inserting Info Into Mysql Table

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Hello, me again.

I have created a little forum - very very basic.

main_forum.php :

Code: [Select]
<?php
$host="localhost"; // Host name
$username="yvonnedp"; // Mysql username
$password="yvonne"; // Mysql password
$db_name="forum"; // Database name
$tbl_name="forum_question"; // Table name

// Connect to server and select databse.
mysql_connect('localhost', 'yvonnedp', 'yvonne')or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");

$sql="SELECT * FROM $tbl_name ORDER BY id";
// OREDER BY id DESC is order result by descending
$result=mysql_query($sql);
?>
<table width="90%" border="0" align="center" cellpadding="3" cellspacing="1" bgcolor="#CCCCCC">
<tr>
<td width="6%" align="center" bgcolor="#E6E6E6"><strong>#</strong></td>
<td width="53%" align="center" bgcolor="#E6E6E6"><strong>Topic</strong></td>
<td width="15%" align="center" bgcolor="#E6E6E6"><strong>Views</strong></td>
<td width="13%" align="center" bgcolor="#E6E6E6"><strong>Replies</strong></td>
<td width="13%" align="center" bgcolor="#E6E6E6"><strong>Date/Time</strong></td>
</tr>

<?php
while($rows=mysql_fetch_array($result)){ // Start looping table row
?>
<tr>
<td bgcolor="#FFFFFF"><? echo $rows['id']; ?></td>
<td bgcolor="#FFFFFF"><a href="view_topic.php?id=<? echo $rows['id']; ?>"><? echo $rows['topic']; ?></a><BR></td>
<td align="center" bgcolor="#FFFFFF"><? echo $rows['view']; ?></td>
<td align="center" bgcolor="#FFFFFF"><? echo $rows['reply']; ?></td>
<td align="center" bgcolor="#FFFFFF"><? echo $rows['datetime']; ?></td>
</tr>

<?php
// Exit looping and close connection
}
mysql_close();
?>
create_topic.php :

Code: [Select]
<table width="400" border="0" align="center" cellpadding="0" cellspacing="1" bgcolor="#CCCCCC">
<tr>
<form id="form1" name="form1" method="post" action="add_topic.php">
<td>
<table width="100%" border="0" cellpadding="3" cellspacing="1" bgcolor="#FFFFFF">
<tr>
<td colspan="3" bgcolor="#E6E6E6"><strong>Create New Topic</strong> </td>
</tr>
<tr>
<td width="14%"><strong>Topic</strong></td>
<td width="2%">:</td>
<td width="84%"><input name="topic" type="text" id="topic" size="50" /></td>
</tr>
<tr>
<td valign="top"><strong>Detail</strong></td>
<td valign="top">:</td>
<td><textarea name="detail" cols="50" rows="3" id="detail"></textarea></td>
</tr>
<tr>
<td><strong>Name</strong></td>
<td>:</td>
<td><input name="name" type="text" id="name" size="50" /></td>
</tr>
<tr>
<td> </td>
<td> </td>
<td><input type="submit" name="Submit" value="Submit" /> <input type="reset" name="Submit2" value="Reset" /></td>
</tr>
</table>
</td>
</form>
</tr>
</table>
add_topic.php :

Code: [Select]
<?php
$host="localhost"; // Host name
$username="yvonnedp"; // Mysql username
$password="yvonne"; // Mysql password
$db_name="forum"; // Database name
$tbl_name="forum_question"; // Table name

// Connect to server and select database.
mysql_connect('localhost', 'yvonnedp', 'yvonne')or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");

// get data that sent from form
$topic=$_POST['topic'];
$detail=$_POST['detail'];
$name=$_POST['name'];

$datetime=date("d/m/y h:i:s"); //create date time

$sql="INSERT INTO $tbl_name(topic, detail, name, datetime)VALUES('$topic', '$detail', '$name', '$datetime')";
$result=mysql_query($sql);

if($result){
echo "Successful<BR>";
echo "<a href=main_forum.php>View your topic</a>";

$email_from = "forums@thenewme.co.za";
     $email_to = "info@thenewme.co.za";
     $email_subject = "The New Me - New Forum Topic";

     $email_message = "A new topic has been posted on the Forum!\n";
     $email_message .= "Check it out : http://www.thenewme.co.za/forum/main_forum.php";

     $headers = 'From: '.$email_from."\r\n";
'Reply-To: '.$email_from."\r\n" . 'X-Mailer: PHP/' . phpversion();
  @mail($email_to, $email_subject, $email_message, $headers);

}
else {
echo "ERROR";
}
mysql_close();
?>
add_answer.php :

Code: [Select]
<?php

$host="localhost"; // Host name
$username="yvonnedp"; // Mysql username
$password="yvonne"; // Mysql password
$db_name="forum"; // Database name
$tbl_name="forum_answer"; // Table name

// Connect to server and select databse.
mysql_connect('localhost', 'yvonnedp', 'yvonne')or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");

// Get value of id that sent from hidden field
$id=$_POST['id'];

// Find highest answer number.
$sql="SELECT MAX(a_id) AS Maxa_id FROM $tbl_name WHERE question_id='$id'";
$result=mysql_query($sql);
$rows=mysql_fetch_array($result);

// add + 1 to highest answer number and keep it in variable name "$Max_id". if there no answer yet set it = 1
if ($rows) {
$Max_id = $rows['Maxa_id']+1;
}
else {
$Max_id = 1;
}

// get values that sent from form
$a_name=$_POST['a_name'];
$a_answer=$_POST['a_answer'];

$datetime=date("d/m/y H:i:s"); // create date and time

// Insert answer
$sql2="INSERT INTO $tbl_name(question_id, a_id, a_name, a_answer, a_datetime)VALUES('$id', '$Max_id', '$a_name', '$a_answer', '$datetime')";
$result2=mysql_query($sql2);

if($result2){
echo "Successful<BR>";
echo "<a href='view_topic.php?id=".$id."'>View your answer</a>";

// If added new answer, add value +1 in reply column
$tbl_name2="forum_question";
$sql3="UPDATE $tbl_name2 SET reply='$Max_id' WHERE id='$id'";
$result3=mysql_query($sql3);

$email_from = "forums@thenewme.co.za";
     $email_to = "info@thenewme.co.za";
     $email_subject = "The New Me - New Forum Answer";

     $email_message = "A new answer to a forum topic has been posted on the Forum!\n";
     $email_message .= "Check it out : http://www.thenewme.co.za/forum/main_forum.php";

     $headers = 'From: '.$email_from."\r\n";
'Reply-To: '.$email_from."\r\n" . 'X-Mailer: PHP/' . phpversion();
  @mail($email_to, $email_subject, $email_message, $headers);

}
else {
echo "ERROR";
}

mysql_close();
?>
view_topic :

Code: [Select]
<?php
$tbl_name2="forum_answer"; // Switch to table "forum_answer"

$sql2="SELECT * FROM $tbl_name2 WHERE question_id='$id'";
$result2=mysql_query($sql2);

while($rows=mysql_fetch_array($result2)){
?>
<table width="400" border="0" align="center" cellpadding="0" cellspacing="1" bgcolor="#CCCCCC">
<tr>
<td><table width="100%" border="0" cellpadding="3" cellspacing="1" bgcolor="#FFFFFF">
<tr>
<td bgcolor="#F8F7F1"><strong>ID</strong></td>
<td bgcolor="#F8F7F1">:</td>
<td bgcolor="#F8F7F1"><? echo $rows['a_id']; ?></td>
</tr>
<tr>
<td width="18%" bgcolor="#F8F7F1"><strong>Name</strong></td>
<td width="5%" bgcolor="#F8F7F1">:</td>
<td width="77%" bgcolor="#F8F7F1"><? echo $rows['a_name']; ?></td>
</tr>
<tr>
<td bgcolor="#F8F7F1"><strong>Answer</strong></td>
<td bgcolor="#F8F7F1">:</td>
<td bgcolor="#F8F7F1"><? echo $rows['a_answer']; ?></td>
</tr>
<tr>
<td bgcolor="#F8F7F1"><strong>Date/Time</strong></td>
<td bgcolor="#F8F7F1">:</td>
<td bgcolor="#F8F7F1"><? echo $rows['a_datetime']; ?></td>
</tr>
</table></td>
</tr>
</table><br>

<?
}

$sql3="SELECT view FROM $tbl_name WHERE id='$id'";
$result3=mysql_query($sql3);

$rows=mysql_fetch_array($result3);
$view=$rows['view'];

// if have no counter value set counter = 1
if(empty($view)){
$view=1;
$sql4="INSERT INTO $tbl_name(view) VALUES('$view') WHERE id='$id'";
$result4=mysql_query($sql4);
}

// count more value
$addview=$view+1;
$sql5="update $tbl_name set view='$addview' WHERE id='$id'";
$result5=mysql_query($sql5);

mysql_close();
?>
the problem is. With add_answer, I am not able to insert an answer. This has worked for the very first question, but all subsequent questions, I am unable to add answers. It keeps telling me ERROR.

This is where the live forum is : http://www.thenewme.co.za/forum/main_forum.php

Can it be an issue with the field that gets autmatically incremented? i have echo'd the data which get entered in when someone types in an answer, and everything is perfect!

Can anyone help me?

Help In Php Mysql Insert Into Query

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Hello every body,, i want to create a new database (auto generate duty assigned using form) in PHP and mysql.. i have four input fields:- Name, Subject, Class & weekly lectures now i want to insert name,subject & class into database, when i insert number of weekly lecture in field four.. insert automaticlly in database multiple time which i write in field four (weekly lecture) database: table structure i have already is::: id-name-subject-class-period-monday-tuesday-wednesday-thursday-friday-saturday anybody please help me to create this database or just insert query ....

Mysql Insert Query

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Hello Friends,

Now i m stuck with simple insert query .

here's the code

Code: [Select]
<?php
include_once("conf.php");

$firstName=$_POST['fname'];
$lastName=$_POST['lname'];
$email=$_POST['email'];
$dob=$_POST['dob'];
$password=$_POST['pass'];

$fname= stripslashes($firstName);
$lname=stripslashes($lastName);

$mail= mysql_real_escape_string($email);
$password= mysql_real_escape_string($password);

mysql_select_db('site');
$statement="Insert into Accounts(Name,lastName,emailId,DOB,password) VALUES($fname,$lname,$mail,$dob,$password)";

$query=mysql_real_escape_string($statement);

mysql_query($query) or die("Cannot save data:</br> ".mysql_error());
echo "Data Saved Successfully";

?>

Now please explain me why i m getting the following error

Cannot save data:
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 '@gmail.com,1988/12/20,password)' at line 1

Any help will be highly appreciated!

Mysql Query Insert In While Loop

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Hi,

Here is an extract of my while loop

while ($personquery = mysql_fetch_array($personfetch, MYSQL_ASSOC))
          {

          echo "$personfetch[first_name]";
}

This will list each person now I want to do an INSERT to the database on every single member that is being listed I can only get it to INSERT for the one.

Thanks

Mysql Multi-query Insert

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Can't Insert Array Into Mysql Query

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I have tried many ways of storing an array of data that comes from a multiple selection form into a mysql table, including serialization. I am currently trying to do it by forming a string, and using the string in the insert query, but keep getting errors, or just nothing happening. I am not sure what I am doing wrong.

Code: [Select]
$categoryString = array();

if ($categoryArray){
foreach ($categoryArray as $category){
$categoryString[] = $category.'<br />';}
}
$categoryquery= "INSERT INTO categoryname VALUES('$categoryString')";

mysql_query($categoryquery);
Thank you for your time as always.

Mysql Question: Loading Database Info From More Than One Table.

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well I know the standard way of retrieving mysql data was through the following codes:

Code: [Select]
$query = "SELECT * FROM {$tablename} WHERE columnmame = '{$var}'";
$result = mysql_query($query);
$row = mysql_fetch_array($result);

This will return all properties inside a table row by an associative array indexed by column names. I am, however, wondering if there is an easier way to retrieve database info from more than one table. For now, what I am doing is:

Code: [Select]
$result = mysql_query( "SELECT * FROM {$tablename} WHERE columnmame = '{$var}'");
$row = mysql_fetch_array($result);

$result2 = mysql_query( "SELECT * FROM {$tablename2} WHERE columnmame2 = '{$var2}'");
$row2 = mysql_fetch_array($result2);

which is a bit tedious and can cause problems when two or more coders work on the same project(it will be difficult to tell what is $row1, $row2 and $row3...). Is there away to write a simpler code than the one above? I mean, if it is possible to run mysql_fetch_array only once and retrieve database info from multiple tables?

Pull Info From One Mysql Table Based On Another's Result

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Hi all,

I have a number of tables within a mysql database and I'm stuck on how to pull information from one based on the results being displayed by another.

The site shows items for sale where by all the product information is held in tablea, when an individual item is being displayed I need to be able to pull a manufacturer profile from tableb based on the manufacturer name stored in tablea column profile. The manufacturer name is already being called into the page by existing code as the site has always shown the manufacturer name, I just can't work out out to pull the new profile information in as well from the new table.

Any help appreciated

Iain

Mysql Insert Query Inside A Php For-loop

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I am really just a beginner in PHP but I've got a task to create a simple hotel booking system by using PHP and MySQL. It works in several steps, described in separate .php files. All of them seem to work fine, except the last one. In this last file I want to add all the data into a MySQL database table. I would like to do this by creating a record for each date, a guest is spending in a hotel (so that when booking, another loop checks whether the room is not occupied yet for that date). I decided to use a for-loop for that, but it doesn't seem to be working (the records just don't get inserted into the table). I checked all the variables by echo-ing them and they are all right. I also tried the query without the loop and it works as well. So I guess there is some other problem. This particular .php file looks as follow:
<?php
session_start();
include('db_config.php');
    $name=($_POST['name']);
    $telephone=addslashes($_POST['telephone']);
    $email=addslashes($_POST['email']);
    $code=md5($email.time());
    $checkout_date=$_SESSION['checkout'];
    $checkin_unix=$_SESSION['checkin_unix'];
    $checkout_unix=$_SESSION['checkout_unix'];
    $roomtype=$_SESSION['roomtype'];
    $checkin_date=$_SESSION['checkin'];
    for ($stay_date=$checkin_unix; ; $stay_date=$stay_date+24*60*60) {
      if ($stay_date=$checkout_unix){
      break;
      }
      mysql_query("INSERT INTO reservations VALUES(
        '',
        '$roomtype',
        'pending',
        'date ('d/m/Y', $stay_date)',
        '$stay_date',
        '$checkout_date',
        '$name',
        '$telephone',
        '$email',
        '$code'
        )");

    }
?>

I would appreciate any kind of help, any idea.

Post Variable, Do Query And Insert Into Table Same Php

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hi .. hit little snag Ive got html <form name="formx" method="POST" action="connect.php" class='ajaxform'> and about 4 varibles in that form .. now how to properly query table users so i can find out which prefix correspond with that user Insert into table "upis" in this case that prefix with 3 variables.. everything in single,.. in this case connect.php file I know how to do QUERY and INSERT INTO but i don't know how to connect with 2 different tbl or db and do the query and input of data.. every time I put 2 of them together .. never works out.. this is example.. i dont know is this right way to do it ..

How To Insert Into A New Table Everytime The Query Is Implemented

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i want a situation where, when i upload something into my database, it creates a new table, everytime, i saw this $forum = "CREATE TABLE IF NOT EXISTS for_".$id." (

How Do I Take Data From A Select Query And Insert Into Another Table?

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I have tried several attempts and can get data to select and using echo display it, however, I need to take this data and insert it into the database in a separate table. I have the following which does hafl the job, can I get some pointers on the rest. I have looked everywhere and not found a solution, at all.

// Selects the data I need
<?php
mysql_connect("PRIVATE INFO","PRIVATE INFO","PRIVATE INFO") or
die("Could not connect: " . mysql_error());

mysql_select_db("wpdb");

$result = mysql_query("SELECT ID FROM wp_posts

WHERE post_title LIKE '%future%'
AND post_status = 'publish'
OR post_title LIKE '%option%'
AND post_content LIKE '%fundamental%'
AND post_status = 'publish'

   ORDER BY post_date DESC");

while ($row = mysql_fetch_array($result, MYSQL_ASSOC))
{   echo $row['ID'];
   echo "<br />";};

?>

Thought maybe something like the following would work, but am at a loss:
INSERT INTO wp_term_relationships (object_id, term_taxonomy_id, term_order)
SELECT ID FROM wp_posts

WHERE post_title LIKE '%future%'
AND post_status = 'publish'
OR post_title LIKE '%option%'
AND post_content LIKE '%fundamental%'
AND post_status = 'publish'

ORDER BY post_date DESC

while ($row = mysql_fetch_array($result, MYSQL_ASSOC))

VALUES
($row['ID'], '25', '0')

Ajax Redirect Doesn't Work After Php Insert Mysql Query

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I have created a registration page to access my website. After the user registrate himself should appear an alert saying that the registration was OK and a redirect to main.php page... however for some reason if I create an insert statement the alert and the redirect don't appear... If I remove the insert the alert and the redirect works... why?

This is part of the code of my 3 files:

registration.php (ajax call)

$('#submit').click(function() {
var username2 = $('#uname2').val();
var password2 = $('#psw2').val();
$.ajax({
url: 'ajax/response.php',
type: 'GET',
data: {username2 : username2, password2: password2},
success: function(data) {
if(data === 'correct') {
alert("Username and Password have been created!"); //don' work with the insert
location.replace("main.php"); //don' work with the insert
} else {
alert("Username or password are not correct... please register yourself!");
}
}
});
});

response.php (answer to ajax call)

if(isset($_GET['username2']) && isset($_GET['password2'])) {
$username2 = $_GET['username2'];
$password2 = $_GET['password2'];
if (checkUser($pdo, $username2) === true) {
echo 'duplicate';
} else {
insertUserPwd($pdo, $username2, $password2); //including this line the redirect and the alert doesn't work... the insert is OK
echo 'correct';
}
}

data_access.php (the function works but doesn't permit alert and redirect to appear)

function insertUserPwd(PDO $pdo, $usr, $pwd)
{
$data = [
'id' => '',
'user' => $usr,
'password' => $pwd
];
$sql = "INSERT INTO users (id, user, password) VALUES (:id, :user, :password)";
$stmt= $pdo->prepare($sql);
$stmt->execute($data);
}

Can someone help me to fix the code?

Insert Into Mysql Table

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Im trying to insert some values automatically into a table once the form loads, but Im getting an error. Here is the code

Code: [Select]
<?php
$aid = $_GET['aid'];
$sd = $_GET['sd'];

?>

<style>
#message {margin:20px; padding:20px; display:block; background:#cccccc; color:#cc0000;}
</style>

<div id="message">Your notification has been submitted.</div>

<div style="text-align:center ">
<?php
$connection = mysql_connect("localhost",
        "username",
        "password");

    mysql_select_db("articles", $connection);

$query="INSERT INTO broken_links (articleid, article) VALUES ('$aid', '$sp')";
if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "Submitted";
mysql_close($con)
?>

<table border="0" cellpadding="3" cellspacing="3" style="margin:0 auto;" >

<input type="submit" id="Login" value="  Thank you. Please press to close  " onclick="tb_remove()"></td>
</tr>
</table>
</div>

Any help will be appreciated

Can't Insert Into Mysql Table

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Hi, The following code was written by someone else. It allows me to upload images to a directory while saving image name in the mysql table. I also want the code to allow me save other data (surname, first name) along with the image name into the table, but my try is not working, only the images get uploaded. What am I missing here?

if(isset($_POST['upload']))
{

$path=$path.$_FILES['file_upload']['name'];

if(move_uploaded_file($_FILES['file_upload']['tmp_name'],$path))
{
echo " ".basename($_FILES['file_upload']['name'])." has been uploaded<br/>";
echo '<img src="gallery/'.$_FILES['file_upload']['name'].'" width="48" height="48"/>';
$img=$_FILES['file_upload']['name'];
$query="insert into imgtables (fname,imgurl,date) values('$fname',STR_TO_DATE('$dateofbirth','%d-%m-%y'),'$img',now())";
if($sp->query($query)){
echo "<br/>Inserted to DB also"; 
}else{
echo "Error <br/>".$sp->error; 
}
}
else
{
echo "There is an error,please retry or check path";
}
}

?>

joseph

Mysql Insert Into And Create Table

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Hey yall!
I'm working on a new site idea and I've run across a problem that I know is simple enough but I'm stumped. It's in the signup form
What I want to do is insert the new user into the 'Login' table and get the user id that was just created and use that to create a table with the user id in the name.

Here is what I have:
// now we insert user into 'Login' table

mysql_real_escape_string($insert = "INSERT INTO `Login` (`UID`, `pass`, `HR`, `mail`, `FullName`) VALUES ('{$_POST['username']}', '{$_POST['pass']}', '{$_POST['pass2']}', '{$_POST['e-mail']}', '{$_POST['FullName']}')");

mysql_query($insert) or die( 'Query string: ' . $insert . '<br />Produced an error: ' . mysql_error() . '<br />' );

$error="Thank you, you have been registered.";

setcookie('Errors', $error, time()+20);

// Get user ID

mysql_real_escape_string($checkID = "SELECT * FROM Login WHERE `mail` = '{$_POST['e-mail']}'");

while ($checkIDdata = mysql_fetch_assoc($checkID)) {

$userID = $checkIDdata;

// now we create table 'Transactions" for the user
mysql_real_escape_string($create = "CREATE TABLE  `financewatsonn`.`Transactions_{$userID}` (
`ID` INT( 20 ) NOT NULL AUTO_INCREMENT COMMENT  'Transaction ID',
`name` VARCHAR( 50 ) NOT NULL COMMENT  'Name/Location',
`amount` VARCHAR( 50 ) NOT NULL COMMENT  'Amount',
`date` VARCHAR( 50 ) NOT NULL COMMENT  'Date',
`category` VARCHAR( 50 ) DEFAULT NULL COMMENT  'Category',
`delete` INT( 1 ) NOT NULL DEFAULT  '0',
UNIQUE KEY  `ID` (  `ID` )
) ENGINE = MYISAM DEFAULT CHARSET = utf8 COMMENT =  'User ID {$userID}'");

mysql_query($create) or die( 'Query string: ' . $create . '<br />Produced an error: ' . mysql_error() . '<br />' );
}

I know in 'Get user ID' that I need to get the ID but I'm not sure how to get that information.

i get the error Quote

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource

at 166 which is the while line under Get user ID

Insert Calculated Value To Mysql Table

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Hi to All,

I am performing calculation which i echo it in the input form to insert it to DB table
The calculation works fine when i submit the form but the it does not correct calculation to the DB.
it seems that the isert to database is done before the calculation and I can't figure a way around it.

Because i'm submitting to the same page, the calculation populate in the input form correctly but it insert Zero to the database table instead of correct calculation populated in the input field to DB table

You could see from the line 114 to 116 that, i'm performing some calculations and echo it at line 128 and line 128 in the input form value field.

Please any help on how to do this...All the function in the code is in another file and it works fine...so the only program is that the calculated value is not inserted as expected

Here is the code

<?php
include 'core/initForMainLogPage.php';


if(isset($_GET['empId']) && !empty($_GET['empId'])){

//delete employee here
$empId=$_GET['empId'];
grabEmpId($empId);
}
?>

<?php
if(logged_in()){
$data=user_dataManager('username');
$usernameData=$data['username'];
}else{
header('Location: index.php');
}
?>
<?php include 'includes/adminHeadAll.php';?>

<header>
<?php include 'includes/managerMenu.php';?>  
</header>
<div class="container">
<br/>
<h3>Pay Employee</h3>
<?php
$error=array();
$errorAll='';
$leave="";
if(isset($_POST['empId']) && isset($_POST['name']) && isset($_POST['date']) && isset($_POST['basicSalary'])
&& isset($_POST['leave']) && isset($_POST['salaryPerDay']) && isset($_POST['leaveDeduct']) 
&& isset($_POST['netSalary'])){
$empId=htmlentities(mysql_real_escape_string($_POST['empId']));
$name=htmlentities(mysql_real_escape_string($_POST['name']));
$date=htmlentities(mysql_real_escape_string($_POST['date']));
$basicSalary=htmlentities(mysql_real_escape_string($_POST['basicSalary']));
$leave=htmlentities(mysql_real_escape_string($_POST['leave']));
$salaryPerDay=htmlentities(mysql_real_escape_string($_POST['salaryPerDay']));
$leaveDeduct=htmlentities(mysql_real_escape_string($_POST['leaveDeduct']));
$netSalary=htmlentities(mysql_real_escape_string($_POST['netSalary']));



//checking for the validity of data entered
if(empty($leave) || empty($date)){
$error[]='Pleave leave or date field is empty.';
}else{

if(preg_match('/[0-9]/',$leave)==false){
$error[]='Leave should only contain numbers'; 
}

if(empId($empId)===false){
$error[]="This employee is not recoganize by the system and can not be paid,he may need to register first.";
}

}

if(!empty($error)){
$errorAll= '<div class="error"><ul><li>'.implode('</li><li>',$error).'</li></ul></div>';


}else{
//this funciton insert into database
payrollData($name,$empId,$date,$basicSalary,$leave,$salaryPerDay,$leaveDeduct,$netSalary);
echo '<p class="pa">Payment made successfully.&nbsp;&nbsp;&nbsp;<a href="employees-salary-report.php">See Payment Records</a></p>';
}


}//end isset


?>

<div class="tableWrap">
<form action="" method="post" >
<div class="styletable"><table cellpadding="" cellspacing="" border="0">
<?php
$query=mysql_query("SELECT 
			  empId,name,level,company.compId,company.levelOne,company.levelTwo,
			  company.levelThree,company.levelFour,company.levelFive
			   FROM employee JOIN company ON company.compId=1 WHERE empId='$empId' LIMIT 1");


while($row=mysql_fetch_array($query)){
  $empId=$row['empId'];
  $name=$row['name'];
  $levelEmp=$row['level'];
  $levelOne=$row['levelOne'];
  $levelTwo=$row['levelTwo'];
  $levelThree=$row['levelThree'];
  $levelFour=$row['levelFour'];
  $levelFive=$row['levelFive'];
  
  if($levelEmp==1){
	  $levelPay=$levelOne;
  }elseif($levelEmp==2){
	  $levelPay=$levelTwo;
  }elseif($levelEmp==3){
	  $levelPay=$levelThree;
  }elseif($levelEmp==4){
	  $levelPay=$levelFour;
  }elseif($levelEmp==5){
	  $levelPay=$levelFive;
  }
  


//making calculations here

$basicSalary=$levelPay * 30;
$leaveDeduct=$leave * $levelPay;
$netSalary=$basicSalary - $leaveDeduct;
}


?>
<tr><td>Employee ID: </td><td><input type="text" name="empId" readonly="readonly" value="<?php if(isset($empId)){echo $empId;}?>"></td></tr>
<tr><td>Employee: </td><td><input type="text" name="name" readonly="readonly" value="<?php if(isset($name)){ echo $name;}?>"></td></tr>
<tr><td>Date: </td><td><input type="text" id="Date" class="picker" name="date"></td></tr>

<tr><td> Basic Salary: </td><td><input type="text" name="basicSalary" readonly="readonly" value="<?php echo $basicSalary;?>"></td></tr>
<tr><td> No. Of Absent: </td><td><input type="text" name="leave" class="input" value=""></td></tr>
<tr><td> Salary Per Day:</td><td><input type="text" name="salaryPerDay" readonly="readonly" value="<?php echo $levelPay;?>"></td></tr>
<tr><td> Deduction For Absentee:</td><td><input type="text" name="leaveDeduct" readonly="readonly" value="<?php echo $leaveDeduct;?>"></td></tr>
<tr><td> Net Salary:</td><td><input type="text" name="netSalary" readonly="readonly" value="<?php echo $netSalary;?>"></td></tr>
<tr><td> </td><td><input type="submit" value="Submit Pay" class="submit" name="pay"></td></tr>
</table></div>

</form>
<?php



?>
</div>
<br />
<?php echo $errorAll; ?>
<p>Manage the monthly salary details of your employee along with the allowances, deductions, etc. by just entering their leave</p>
</div> 
<?php include 'includes/footerAll.php';?>  

<script type="text/javascript" src="js/jquery.js"></script>
<script type="text/javascript" src="js/jquery-ui.js"></script>
<script type="text/javascript" src="js/ui.js"></script>
</body>
</html>

Mysql Insert Into Table Values

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What gets put in for the values is my problem. I have tried every way I can think of but I get errors in the code or I don't get the row inserted.

Can anyone give me a code sample using $ variables, preferably a multi-line list? I would have to fix up my code to show you what I am trying as it is in between tries now.

I have to be doing something wrong and may have to put together a simple test script but a working example would be better. Everything I look at on the web shows "xxxxx xxxx" examples not $variables.

Thanks in advance.