insert...why won't this work? :(

whoisKeel

I just want to insert a new row, taking things one at a time

i can logon to my phpAdmin on web and insert new rows and i am copying the code they give me there even, but still, this code will not insert new rows...i can even create tables within html no problem...what gives?

btw, it doesn't give me a failed to query message...it just doesn't put the data in the db...

<?php

$connection = mysql_connect("localhost","myUsername","myPassword") or die("Couldn't connect to server.");
$db = mysql_select_db("myDB", $connection) or die("Couldn't connect to server.");
$whatever = "workdammit";

$sql = "INSERT INTO $workdammit ( ID , firstName , lastName , age ) VALUES ( \'003\', \'Zeus\', \'McKenny\', \'23\' ); ";

mysql_query( $sql, $connection );
$sql = "SELECT * FROM $whatever;";
$sql_result = mysql_query($sql, $connection) or die("Couldn't execute query.");
?>

tserbis

$sql = "INSERT INTO $whatever ( ID , firstName , lastName , age ) VALUES ( '003', 'Zeus', 'McKenny', '23' )";

whoisKeel

yes i realized that after i had posted 😉

but that still did not fix my problem

but this did:

<?php
$connection = mysql_connect("localhost","blah","blah") or die("Couldn't connect to server.");
$whatever = "newtable";
mysql_select_db("myDB");
mysql_query("INSERT INTO $whatever (ID, firstName, lastName, age) values ('003', 'Zeus', 'McKenny', '25')");
?>

it seems that calling "mysql_select_db("myDB");" directly fixed it...i don't understand this as every tutorial on the web and my book do not do it this way.... anyone know why?

csousley

You had another problem '$workdammit ' in the first thread is never defined. 'workdammit' is the value of a variable 'whatever', but '$workdammit' doesn't exist.

Whenever I'm having a hard time with a query, I'll set the query into a variable and echo the query variable on the page to make sure all the variables in the query are there...Make Sense?

$whatevvvvvvvvver = 'mytable';
$query = 'select * from $whatever';
echo $query;

when the query echo'd it would show

select * from

and I would know there was something wrong with '$whatever'.

kula

Hi,
below code may help for your question

<?php

$connection = mysql_connect("localhost","myUsername","myPassword") or die("Couldn't connect to server.");

if(! mysql_select_db("myDB")
{
echo "Couldn't connect to server.";
exit;
}
$whatever = "workdammit";

$sql = "INSERT INTO $workdammit ( ID,firstName, lastName,age)
VALUES ( '003', 'Zeus', 'McKenny', '23'); ";

$res=mysql_query( $sql, $connection );

if (!$res) {
print "ERROR: <P>$sql</P>";
print mysql_error();
exit;
}

$sql = "SELECT * FROM $whatever;";
$sql_result = mysql_query($sql, $connection)

if (!$sql_res) {
print "ERROR: <P>$sql</P>";
print mysql_error();
exit;
}
?>