php and mysql help

I need to make a web page in php and search a my SQL database. For data in three the field. And right back to the database.

If you don't post any code for people to look through then your threads are not going to be answered.

Do you know who to create a webpage?

yes html

Code:

---

------------ 1- Filename: config.php ------------ ------------ 1- Filename: search.php ------------
Search Word: 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Program Files\xampp\htdocs\web\hi.php on line 31

---

Code:

---

------------
1- Filename: config.php
------------
<?php
//define your database informations
$dbhost="localhost";
$dbusername="root";
$dbpassword="crocker";
$dbname="databasename";
?>


------------
1- Filename: search.php
------------
<?php
include("config.php");
        // connect to db
  $db = mysql_connect($dbhost, $dbusername, $dbpassword);
    mysql_select_db($dbname,$db);
  // The form:
?>
<form method="POST" action="">
Search Word: <input type="text" name="query">
<input type="SUBMIT" value="Search!">
</form>

<?
// PHP Search Script
$result = mysql_query("select * from Table_Name WHERE column1 LIKE '%$query%' OR column2 LIKE '%$query%'", $db);
    while(list($column1, $column2)=mysql_fetch_array($result)){
          echo "Result: $column1, $column2 <br />";

}
?>

---

You have not declared your variable "query".

Also you have not set a value for "action" in your form.

You could use:

PHP Code:

---

action="$PHP_SELF"
or
action="http://www.yoursite.com/somepage.php" 


---

did not work for me.

Quote:

---

Originally Posted by lintz

You have not declared your variable "query".

---

Where have you declared this variable?

I got the form http://www.phpfreaks.com/quickcode/B...iglight=search

Have you changed the database variables to relate to your database?

Also i doubt you have a table called "Table_Name" and fields called "column1" and "column2" :confused:

EDIT: You have changed the db variables.....my mistake.

no
table called "name"
fields called "FirstName" and LastName

OK. Replace the name of the table and field names in the example with yours.

I am new to php.
how?

PHP Code:

---

$result = mysql_query("select * from name WHERE FirstName LIKE '%$query%' OR LastName LIKE '%$query%'", $db); 


---

Code:

---

------------
1- Filename: config.php
------------
<?php
//define your database informations
$dbhost="localhost";
$dbusername="root";
$dbpassword="crocker";
$dbname="school";
?>


------------
1- Filename: search.php
------------
<?php
include("config.php");
        // connect to db
  $db = mysql_connect($dbhost, $dbusername, $dbpassword);
    mysql_select_db($dbname,$db);
  // The form:
?>
<form method="POST" action="brb.no-ip.info">
Search Word: <input type="text" name="query">
<input type="SUBMIT" value="Search!">
</form>

<?
// PHP Search Script
$result = mysql_query("select * from Name WHERE FirstName LIKE '%$query%' OR LastName LIKE '%$query%'", $db);
    while(list($FirstName, $LastName)=mysql_fetch_array($result)){
          echo "Result: $FirstName, $LastName <br />";

---

to test it goto http://bob5731.no-ip.info/web/search.php

Object not found! error.

It will have to do with action="brb.no-ip.info". You need to specify a location on your site for the form resutls to goto. eg. action="brb.no-ip.info/resutls.php

Object not found!
The requested URL was not found on this server. The link on the referring page seems to be wrong or outdated. Please inform the author of that page about the error.

If you think this is a server error, please contact the webmaster.

Error 404
bob5731.no-ip.info
11/07/05 22:36:44
Apache/2.0.54 (Win32) mod_ssl/2.0.54 OpenSSL/0.9.8 PHP/5.0.5 mod_autoindex_color

Give me a couple of mins and I'll have a working example.

ok!!!

Here you go. This works on my machine :)

This goes in your search.php file. Leave config.php alone if all variables are correct.

I hope you rate my post after this :D

PHP Code:

---

<?php
include("config.php");
$query = $_POST['query'];
         // connect to db
   $db = mysql_connect($dbhost, $dbusername, $dbpassword);
    mysql_select_db($dbname,$db) or die('error connection to DB '.mysql_error());
   // The form:
if (!isset($_POST['submit']) ) { // if form hasn't been submitted to itself then show form

?>
<form method="POST" action="<?php echo $PHP_SELF?>">
Search Word: <input type="text" name="query">
<input type="SUBMIT" name="submit" value="Search!">
</form>

<?

}

else {

// PHP Search Script
$result = mysql_query("select * from Name WHERE FirstName LIKE '%$query%' OR LastName LIKE '%$query%'", $db);
     while(list($FirstName, $LastName)=mysql_fetch_array($result)){
           echo "Result: $FirstName, $LastName <br />";
           }
           
}
?>

---

/*$dbhost="localhost"; $dbusername="root"; $dbpassword="crocker"; $dbname="school"; */
Warning: mysql_connect() [function.mysql-connect]: Access denied for user 'ODBC'@'localhost' (using password: NO) in C:\Program Files\xampp\htdocs\web\search.php on line 12

Warning: mysql_select_db(): supplied argument is not a valid MySQL-Link resource in C:\Program Files\xampp\htdocs\web\search.php on line 13
error connection to DB Access denied for user 'ODBC'@'localhost' (using password: NO)

have you got php, mysql and apache installed on your machine?

It works for me now.
Now how do add data to the mySQL database.

PHP Code:

---

$add_item = mysql_query("INSERT into name VALUES('Will', 'Perdue')", $db);
$add_item = mysql_query("INSERT into name VALUES('John', 'Paxson')", $db);
//etc... 


---

Code:

---

<?php
//define your database informations
$dbhost="localhost";
$dbusername="root";
$dbpassword="crocker";
$dbname="school";
?>
<?php
include("config.php");
$query = $_POST['query'];
        // connect to db
  $db = mysql_connect($dbhost, $dbusername, $dbpassword);
    mysql_select_db($dbname,$db) or die('error connection to DB '.mysql_error());
  // The form:
if (!isset($_POST['submit']) ) { // if form hasn't been submitted to itself then show form

?>
<form method="POST" action="<?php echo $PHP_SELF?>">
Search Word: <input type="text" name="query">
<input type="SUBMIT" name="submit" value="Search!">
</form>

<?

}

else {

// PHP Search Script
$result = mysql_query("select * from Name WHERE FirstName LIKE '%$query%' OR LastName LIKE '%$query%'", $db);
    while(list($FirstName, $LastName)=mysql_fetch_array($result)){
          echo "Result: $FirstName, $LastName <br />";
          }
$add_item = mysql_query("INSERT into name VALUES('bob', 'Perdue')", $db);
$add_item = mysql_query("INSERT into name VALUES('brandon', 'Paxson')", $db);

}
?>

---

It will be for gift card.

http://bob5731.no-ip.info/web/untitled.BMP

So whats wrong with it?

I have no code.

I've never made a gift card, so, neither do I. ave a go at writing it yourself and when you get a problem, post here ;)

It will be gift card db.

Use my example and then modify it to your needs for the gift card.

Is not work for me.

Code:

---

$add_item = mysql_query("INSERT into name VALUES('Will', 'Perdue')", $db);
$add_item = mysql_query("INSERT into name VALUES('John', 'Paxson')", $db);

---

Code:

---

<?php
//define your database informations
$dbhost="localhost";
$dbusername="root";
$dbpassword="crocker";
$dbname="school";
?>
<?php
include("config.php");
$query = $_POST['query'];
        // connect to db
  $db = mysql_connect($dbhost, $dbusername, $dbpassword);
    mysql_select_db($dbname,$db) or die('error connection to DB '.mysql_error());
  // The form:
if (!isset($_POST['submit']) ) { // if form hasn't been submitted to itself then show form

?>
<form method="POST" action="resutls.php"<?php echo $PHP_SELF?>">
Search Word: <input type="text" name="query">
<input type="SUBMIT" name="submit" value="Search!">
</form>

<?

}

else {

// PHP Search Script
$result = mysql_query("select * from Name WHERE FirstName LIKE '%$query%' OR LastName LIKE '%$query%'", $db);
    while(list($FirstName, $LastName)=mysql_fetch_array($result)){
          echo "Result: $FirstName, $LastName <br />";
          }
$add_item = mysql_query("INSERT into name VALUES('bob', '$FirstName')", $db);
$add_item = mysql_query("INSERT into name VALUES('brandon', '$FirstName')", $db);

}
?>

---

OK, as a general rule when some code you're working with doesn't do what you want it returns an error. To help people help you fix the error you should post the code you're working with along with any error messages.

Please post the error message(s) when you attempt to add records to your table.

I get no errors.
It returns a blank page.

http://bob5731.no-ip.info/web/untitled1.BMP

You didn't say you had more then 2 fields in your table :(

You need to add an item for each field in your table.

eg.

PHP Code:

---

$add_item = mysql_query("INSERT into name VALUES('', 'John', '', 'Paxson', '', '', '', '', '', '', '', )", $db)  or die(''.mysql_error());
//mysql_error() will tell you where your error is (if there is one) 


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how to display basic search results mysql using php.

Look at post #20

wood help I gave you the files.