sparky

February 21, 2014

February 6, 2014

I have an issue as to where I am trying to save long amounts of data into a database, and nothing that I try seems to work.... The field $detail when I have more than a coupld lines of data just does not write the data... Note code below:

Viewing the $SQL Data all shows correct... Just does not write it to the Database. I have tried changing to blob and other selections in the MYSQL, but not seeing what I am doing wrong. thanks in advance for any advice.

John

Database:

----------------------

-- phpMyAdmin SQL Dump
-- version 3.5.0
-- http://www.phpmyadmin.net
--
-- Host: localhost
-- Generation Time: Feb 06, 2014 at 09:30 AM
-- Server version: 5.5.23
-- PHP Version: 5.3.13

SET SQL_MODE="NO_AUTO_VALUE_ON_ZERO";
SET time_zone = "+00:00";

/*!40101 SET @OLD_CHARACTER_SET_CLIENT=@@CHARACTER_SET_CLIENT */;
/*!40101 SET @OLD_CHARACTER_SET_RESULTS=@@CHARACTER_SET_RESULTS */;
/*!40101 SET @OLD_COLLATION_CONNECTION=@@COLLATION_CONNECTION */;
/*!40101 SET NAMES utf8 */;

--
-- Database: `elsa`
--

-- --------------------------------------------------------

--
-- Table structure for table `timeline2`
--

CREATE TABLE IF NOT EXISTS `timeline` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`mo` text,
`mon` text,
`da` text,
`yr` text,
`hdr` text,
`detail` longtext,
PRIMARY KEY (`id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 COMMENT='Elsa Time Line' AUTO_INCREMENT=4 ;

/*!40101 SET CHARACTER_SET_CLIENT=@OLD_CHARACTER_SET_CLIENT */;
/*!40101 SET CHARACTER_SET_RESULTS=@OLD_CHARACTER_SET_RESULTS */;
/*!40101 SET COLLATION_CONNECTION=@OLD_COLLATION_CONNECTION */;

----------------------

Insert Code:

----------------------

<?php
// open the connection
$conn = mysql_connect("localhost", "xxxx", "xxxx");

// pick the database to use

mysql_select_db("elsa",$conn);

// create the SQL statement

$mo=$_POST[mo];
$mon=$_POST[mon];
$da=$_POST[da];
$yr=$_POST[yr];
$hdr=$_POST[hdr];
$detail=$_POST[detail];

$sql = "INSERT INTO timeline2(mo, mon, da, yr, hdr, detail) values ('$mo', '$mon', '$da', '$yr', '$hdr', '$detail')";

echo $sql;
echo "<BR>";

// execute the SQL statement
if (mysql_query($sql, $conn)) {
echo "Record Updated!";
} else {
echo "Something went Wrong";
}

?>

----------------------

Example data that does not work.

----------------------

<P ALIGN=CENTER>
   Elsa's Journey to her new home in Memphis</P>
<CENTER>
   <P ALIGN=CENTER>
    <iframe title="YouTube video player" class="youtube-player" type="text/html" width="640" height="510" src="http://www.youtube.com/embed/EJufDyBse8s" frameborder="0" allowFullScreen></iframe></P>
   </CENTER>
<P ALIGN=CENTER>
    </P>
<P ALIGN=CENTER>
   Elsa at Home</P>
<P ALIGN=CENTER>
   <A HREF="assets/140131/index.htm"><IMG SRC="assets/140131/elsa-013114-01_std.jpg" WIDTH="536" HEIGHT="800" VSPACE="0" HSPACE="0" BORDER="0">
<BR>
Click on IMAGE
</A></P>
<P ALIGN=CENTER>
    </P>
<P ALIGN=CENTER>
   Elsa VS the Video Camera.</P>
<CENTER>
   <P ALIGN=CENTER>
    <iframe title="YouTube video player" class="youtube-player" type="text/html" width="640" height="510" src="http://www.youtube.com/embed/DTWntrEVuUs" frameborder="0" allowFullScreen></iframe></P>
   </CENTER>
<P ALIGN=CENTER>
   A Very long Day for just a little Kitty...</P>
<P ALIGN=CENTER>
   <IMG SRC="assets/140131/Sleeping.jpg" WIDTH="640" HEIGHT="480" VSPACE="0" HSPACE="0" BORDER="0"></P>
<P ALIGN=CENTER>
   She was laying under her cat tree... When I turned off the lights
   (from remote) she got up and went to her cat bed and fell asleep.</P>

December 30, 2010

It has been a while since I have done much programming at all, and where I know I have been through this before I am getting an "Undefined Varible Error"

Where this works coming from a form submission:

echo $_POST["co1"];

echo $_POST["co2"];

echo $_POST["co3"];

This does not work:

$s1 = $_post["co1"];

$s2 = $_post["co2"];

$s3 = $_post["co3"];

What stupid function am I overlooking or doing wrong.

Thanks in advance.

John

October 22, 2003

I need a routine that will change all data that is in a field to something else...

Example: where class=\"abc\", I want to change it to class=\"123\"/

Thanks in advance.

August 13, 2003

Not sure.. I am quite a novice... I tried using replace command and really got lost with it... I finally got the replace command working and it is doing great now. Thanks for the comment. All of them are welcome.

sparky

August 12, 2003

Minor goof on my end... Is writing data, but duplicating the record. I feel that I am pretty close, but not quite there. WHen it duplicates it, it also is incrementing the index counter.

August 12, 2003

REPLACE does exactly the same thing as INSERT INTO (even has the same syntax) except it replaces the old data instead of adding a new line.

You tell the DBMS what fields you are replacing then you need to specify which row you are replacing data with in the REPLACE clause, so:
REPLACE INTO table (field1, field2, field3)

SELECT field1, field2, field3 FROM table WHERE tableID=$tableID
In the case of an on-the-fly operation, use:
REPLACE INTO table (fieldID, field2, field3)

VALUES ($fieldID, $field2, $field3)
Is that better?

I am trying to use:

$sql = \"REPLACE INTO newinv (id, item, dir, fn, price, class, dis, inv) VALUES (\'$_POST[id]\', \'$_POST[item]\', \'$_POST[dir]\' , \'$_POST[fn]\', \'$_POST[price]\', \'$_POST[class]\',\'$_POST[dis]\',\'$_POST[inv]\')\";

Is there a problem with the \'$_POST[id]\', routine... WHen I use this, for some reason it is writing a new record that is blank.

August 12, 2003

You need a SELECT in your REPLACE statement or do the VALUES as a line:

1. REPLACE INTO table ()

VALUES ();

2. REPLACE INTO table ()

SELECT select_clause;

I am not understanding the SELECT clause on this. I am missing the big picture on this for some reason.

August 12, 2003

I feel like I am overlooking so ever minor, but unable to find it. What the problem is, when I edit a record, it either does not save, or it creates a duplicate. Make note, my insert code works perfectly, but my replace code does not. Note both:

works for inserting records

<?php

// open the connection

$conn = mysql_connect(\"localhost\", \"xxxx\", \"xxxx\");

// pick the database to use

mysql_select_db(\"galleria\",$conn);

// create the SQL statement

$sql = \"INSERT INTO newinv values (\'*\', \'$_POST[item]\', \'$_POST[dir]\', \'$_POST[fn]\', \'$_POST[price]\', \'$_POST[class]\', \'$_POST[dis]\', \'$_POST[inv]\')\";

// execute the SQL statement

if (mysql_query($sql, $conn)) {

echo \"Record Updated!\";

} else {

echo \"Something went Wrong\";

}

?>

Does NOT work for Editing/Replacing records

<?php

// open the connection

$conn = mysql_connect(\"localhost\", \"xxxx\", \"xxxx\");

// pick the database to use

mysql_select_db(\"galleria\",$conn);

// create the SQL statement

$sql = \"REPLACE INTO newinv (\'$_POST[id]\', \'$_POST[item]\', \'$_POST[dir]\', \'$_POST[fn]\', \'$_POST[price]\', \'$_POST[class]\', \'$_POST[dis]\', \'$_POST[inv]\') WHERE \'id\'=id\";

// execute the SQL statement

if (mysql_query($sql, $conn)) {

echo \"Record Updated!\";

} else {

echo \"Something went Wrong\";

}

?>

-------------------------

The following is the page following that generates the data to be saved.

<html>

<?php

// open the connection

$conn = mysql_connect(\"localhost\", \"xxxx\", \"xxxx\");

// pick the database to use

mysql_select_db(\"galleria\",$conn);

// create the SQL statement

$ids = $_GET[ \'ids\' ];

$sql = \"SELECT * FROM newinv WHERE id=\'$ids\'\";

// execute the SQL statement

$result = mysql_query($sql, $conn) or die(mysql_error());

//go through each row in the result set and display data

while ($newArray = mysql_fetch_array($result)) {

// give a name to the fields

$id = $newArray[\'id\'];

$item = $newArray[\'item\'];

$dir = $newArray[\'dir\'];

$fn = $newArray[\'fn\'];

$price = $newArray[\'price\'];

$class = $newArray[\'class\'];

$dis = $newArray[\'dis\'];

$inv = $newArray[\'inv\'];

}

?>

<tr>

Item Name

</td>

</td> </tr>

<tr>

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sparky

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