| MySQL Help Post any question relating to MySQL here and hopefully someone can help |

2004-09-08, 08:32 PM
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Junior Member
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Join Date: Sep 2004
Posts: 2
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I'm trying to insert data in a table I named "INFO" from web browser
using Html,PHP, to Domain server supporting MYsqlAdmin.
I am able to connect succefully
The problem is inserting Data (using php)
from variables I already assigned.
$First, $Last, and $Email
I used this to connect:
mysql_connect ("$Cerver, "$User", "$Pw") or
die ("couldn't Connect")
Now I want to insert into table "INFO" "First, Last, Email"
the values $First, $Last, $Email.
I tried this:
INSERT INTO `INFO` (`First` , `Last` , `Email`)
VALUES ( '$First' , '$Last' , '$Email');
And got this:
Parse error: parse error, unexpected T_STRING in (the line where my Insert INFO starts)
Did I miss something?
Thanks!
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2004-09-08, 09:34 PM
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Moderator
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Join Date: May 2004
Location: Portugal
Posts: 143
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Hope you haven't forgot to use mysql_query() to run your query :P
Btw, you don't need to put the ' ´' in the table's name and/or fields.
And... don't forget that tables and fields' names are case sensitive.
Try using this:
Code:
<?php
mysql_query("INSERT INTO INFO (First , Last , Email) VALUES ( '" . $First . "' , '" . $Last . "' , '" . $Email . "')");
// Or....
mysql_query("INSERT INTO INFO VALUES ( '" . $First . "' , '" . $Last . "' , '" . $Email . "')");
?>
__________________
Best Regards,
Gonçalo "GesF" Fontoura
Website : gesf.org
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2004-09-10, 06:58 PM
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Thanks gesf
the error is gone now.
The script takes form fields entered, turns them into php variables and inserts those variables into INFO.
It connects and doesn't display any inserting errors.
The problem
When i log on mysqladmin and browse the INFO database, it shows as nothing being inserted.
The variables do have data on them through $_GET because I echoed them out before I inserted them so i'm pretty sure its not a blank data or "null" issue.
What else have i missed? :huh:
Any closing-query command
or database select needed before inserting into a database?
thanks again
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2004-09-11, 06:35 PM
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Moderator
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Join Date: May 2004
Location: Portugal
Posts: 143
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I see! Use the $_GET and/or $_POST superglobal arrays.
Use $_GET to retrieve the data from the URL, or if it was sent through form with GET method.
Use $_POST to retrieve data sent through form with POST method.
Example:
Code:
<?php
mysql_query("INSERT INTO INFO VALUES ( '" . $_POST['First'] . "' , '" . $_POST['Last'] . "' , '" . $_POST['Email'] . "')");
?>
Cheers
__________________
Best Regards,
Gonçalo "GesF" Fontoura
Website : gesf.org
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2004-09-15, 04:44 PM
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What I want to do
Web form simulation
-----------------------------------------------------------
Enter Your
Last Name: [ ]
First Name:[ ]
Email: [ ]
Then
click
[submit"]
----------------------------------------------------------------------------------
I created a basic web form (HTML) that will ask you to Input your Last Name, First Name and Email address.
as shown above.
Once it recieves the input
I ask it to get the Values inserted by user using $_GET
I assign each of the 3 input fields to 3 different variables.
$Last
$First
$Email
I want to connect to mysqladmin server (webhost) and insert (update) a Table I created called INFO
This is my code
Code:
<?
// Variables = Info collected on HTML webform through GET
$First= $_GET['First'];
$Last= $_GET['Last'];
$Email= $_GET['Email'];
// The Server (Mysqladmin web host godaddy.com ), The User and Pass
$Host= "Host.com";
$User= "Username";
$Pass= "Password";
// Connect to database or "couldn't connect" will be display
mysql_connect("$Host", "$User","$Pass" )
or
die ("Couldn't connect ");
// inserting data into a table called "INFO"
mysql_query("INSERT INTO INFO (First , Last , Email) VALUES ( '" . $First . "' , '" . $Last . "' , '" . $Email . "')");
mysql_close();
?>
Everything works , It recieves, It connects without errors (thanks to your help)
but when i log on to mysqladmin and check if the table "INFO" has any input, it shows as no records.
:blink:
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2004-09-16, 06:18 AM
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Moderator
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Join Date: May 2004
Location: Portugal
Posts: 143
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Hunn, not sure!
Try to use this:
Code:
mysql_query("INSERT INTO INFO (First , Last , Email) VALUES ( '$First' , '$Last' , '$Email')");
Instead of this:
Code:
mysql_query("INSERT INTO INFO (First , Last , Email) VALUES ( '" . $First . "' , '" . $Last . "' , '" . $Email . "')");
In last case try to change the for method to post, as well as the variables:
Code:
$First= $_GET['First'];
$Last= $_GET['Last'];
$Email= $_GET['Email'];
Cheers
__________________
Best Regards,
Gonçalo "GesF" Fontoura
Website : gesf.org
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2004-09-25, 03:26 PM
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This is maybe true, but i don`t see mysql_select_db anywhere, some thing like this :
Code:
$mysql = mysql_select_db($database_dbi, $dbi);
if (!$mysql) {
echo "Cannot select database";
}
$query = "select count(*) from users where
passwd=md5($passwdc);
$result = mysql_query($query);
if(!$result) {
echo "Cannot run query";
}
where :
Code:
$dbi = mysql_connect($hostname_dbi, $username_dbi, $password_dbi) or die(mysql_error());
$database_dbi = "your db";
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