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ok problems entering data into database

ok so i am having problems if you look at the script below you will
see that it the query has 4 values to insert but the actual values only
contain title entry and now() for the date. well i have made the
database and the blog_id is a primary auto interger what ever table
bascly look below the the insert code block to find the code block
that makes the table in the database,
// Define the query.
$query = "INSERT INTO blog_entries (blog_id, title, entry,
date_entered) VALUES ('{$_POST['title']}', '{$_POST['entry']}',
NOW())";

// Execute the query.
if (@mysql_query ($query)) {
print '<p>The blog entry has been added.</p>';
} else {
print "<p>Could add the entry because: <b>" . mysql_error() . "</b>.
The query was $query.</p>";
}

---------------------------------------------------------------------------------------------------------------------------------------------------
// Define the query.
$query = 'CREATE TABLE blog_entries (
blog_id INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,
title VARCHAR(100) NOT NULL,
entry TEXT NOT NULL,
date_entered DATETIME NOT NULL
)';

// Run the query.
if (@mysql_query ($query)) {
print '<p>The table has been created.</p>';
} else {
die ('<p>Could not create the table because: <b>' . mysql_error() .
'</b>.</p><p>The query being run was: ' . $query . '</p>');
}

---------------------------------------------------------------------------------------------------------------------------------------------
>>>>blog_id INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,<<<<
basicly that right above is what the table looks like now though if i
go and run the insert script it says that the collumns dont match the
values ect. but how can i get it to utilize the blog id table?
what do i enter as a value?

Nov 22 '06 #1
4 1629
so many sites so little time wrote:
ok so i am having problems if you look at the script below you will
see that it the query has 4 values to insert but the actual values only
contain title entry and now() for the date. well i have made the
database and the blog_id is a primary auto interger what ever table
bascly look below the the insert code block to find the code block
that makes the table in the database,
// Define the query.
$query = "INSERT INTO blog_entries (blog_id, title, entry,
date_entered) VALUES ('{$_POST['title']}', '{$_POST['entry']}',
NOW())";

// Execute the query.
if (@mysql_query ($query)) {
print '<p>The blog entry has been added.</p>';
} else {
print "<p>Could add the entry because: <b>" . mysql_error() . "</b>.
The query was $query.</p>";
}

---------------------------------------------------------------------------------------------------------------------------------------------------
// Define the query.
$query = 'CREATE TABLE blog_entries (
blog_id INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,
title VARCHAR(100) NOT NULL,
entry TEXT NOT NULL,
date_entered DATETIME NOT NULL
)';

// Run the query.
if (@mysql_query ($query)) {
print '<p>The table has been created.</p>';
} else {
die ('<p>Could not create the table because: <b>' . mysql_error() .
'</b>.</p><p>The query being run was: ' . $query . '</p>');
}

---------------------------------------------------------------------------------------------------------------------------------------------
>>>blog_id INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,<<<<
basicly that right above is what the table looks like now though if i
go and run the insert script it says that the collumns dont match the
values ect. but how can i get it to utilize the blog id table?
what do i enter as a value?
Since blog_id is an auto_incrementing field, you shouldn't explicitly
refer to it. MySQL will handle it, then you can get the auto created
value. Try this...

$query = "INSERT INTO blog_entries (title, entry, date_entered) VALUES
('{$_POST['title']}', '{$_POST['entry']}', NOW())";
mysql_query($query);
printf("Last inserted record has id %d\n", mysql_insert_id());

http://us3.php.net/mysql_insert_id

Nov 22 '06 #2
> ok so i am having problems if you look at the script below you will
>see that it the query has 4 values to insert but the actual values only
contain title entry and now() for the date. well i have made the
database and the blog_id is a primary auto interger what ever table
bascly look below the the insert code block to find the code block
that makes the table in the database,
// Define the query.
$query = "INSERT INTO blog_entries (blog_id, title, entry,
date_entered) VALUES ('{$_POST['title']}', '{$_POST['entry']}',
NOW())";

// Execute the query.
if (@mysql_query ($query)) {
print '<p>The blog entry has been added.</p>';
} else {
print "<p>Could add the entry because: <b>" . mysql_error() . "</b>.
The query was $query.</p>";
}

---------------------------------------------------------------------------------------------------------------------------------------------------
// Define the query.
$query = 'CREATE TABLE blog_entries (
blog_id INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,
title VARCHAR(100) NOT NULL,
entry TEXT NOT NULL,
date_entered DATETIME NOT NULL
)';

// Run the query.
if (@mysql_query ($query)) {
print '<p>The table has been created.</p>';
} else {
die ('<p>Could not create the table because: <b>' . mysql_error() .
'</b>.</p><p>The query being run was: ' . $query . '</p>');
}

---------------------------------------------------------------------------------------------------------------------------------------------
>>>>>blog_id INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,<<<<
basicly that right above is what the table looks like now though if i
go and run the insert script it says that the collumns dont match the
values ect. but how can i get it to utilize the blog id table?
what do i enter as a value?
You have three columns and two values in your insert. You need
the number of columns and the number of values to match.

If you leave the blog_id column out of the insert, it will get
filled in with an auto-increment value. This, I think, is what you
want.

If you leave the blog_id column in the insert, and supply null as
a matching value, it will still get filled in with an auto-increment
value. This, I think, is what you want, done another way.

On the other hand, if you leave the blog_id column in the insert,
and supply a number as a matching value, it will insert that value.
Among other things, this lets you restore tables from a backup
without messing up the auto-increment entries.
Nov 22 '06 #3
yeah i think im gonna insert null i just wat to be able to post text so
that it appears like the content of a web page and then can be edited
but when ive been going and trying to edit it say i say home_id 1 or
what ever and then i tell it to get from row 1 it returns a error
which i tell it to if it didnt get an id

Gordon Burditt wrote:
ok so i am having problems if you look at the script below you will
see that it the query has 4 values to insert but the actual values only
contain title entry and now() for the date. well i have made the
database and the blog_id is a primary auto interger what ever table
bascly look below the the insert code block to find the code block
that makes the table in the database,
// Define the query.
$query = "INSERT INTO blog_entries (blog_id, title, entry,
date_entered) VALUES ('{$_POST['title']}', '{$_POST['entry']}',
NOW())";

// Execute the query.
if (@mysql_query ($query)) {
print '<p>The blog entry has been added.</p>';
} else {
print "<p>Could add the entry because: <b>" . mysql_error() . "</b>.
The query was $query.</p>";
}

---------------------------------------------------------------------------------------------------------------------------------------------------
// Define the query.
$query = 'CREATE TABLE blog_entries (
blog_id INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,
title VARCHAR(100) NOT NULL,
entry TEXT NOT NULL,
date_entered DATETIME NOT NULL
)';

// Run the query.
if (@mysql_query ($query)) {
print '<p>The table has been created.</p>';
} else {
die ('<p>Could not create the table because: <b>' . mysql_error() .
'</b>.</p><p>The query being run was: ' . $query . '</p>');
}

---------------------------------------------------------------------------------------------------------------------------------------------
>>>>blog_id INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,<<<<
basicly that right above is what the table looks like now though if i
go and run the insert script it says that the collumns dont match the
values ect. but how can i get it to utilize the blog id table?
what do i enter as a value?

You have three columns and two values in your insert. You need
the number of columns and the number of values to match.

If you leave the blog_id column out of the insert, it will get
filled in with an auto-increment value. This, I think, is what you
want.

If you leave the blog_id column in the insert, and supply null as
a matching value, it will still get filled in with an auto-increment
value. This, I think, is what you want, done another way.

On the other hand, if you leave the blog_id column in the insert,
and supply a number as a matching value, it will insert that value.
Among other things, this lets you restore tables from a backup
without messing up the auto-increment entries.
Nov 23 '06 #4

so many sites so little time schreef:
yeah i think im gonna insert null i just wat to be able to post text so
that it appears like the content of a web page and then can be edited
but when ive been going and trying to edit it say i say home_id 1 or
what ever and then i tell it to get from row 1 it returns a error
which i tell it to if it didnt get an id
>// Define the query.
>$query = 'CREATE TABLE blog_entries (
> blog_id INT UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,
> title VARCHAR(100) NOT NULL,
> entry TEXT NOT NULL,
> date_entered DATETIME NOT NULL
> )';
Here you write something like this:

INSERT INTO A_TABLE (Column_1, Column_2, Column_3,
Column_4) VALUES (Value_2, Value_3, Column_4)

as mentioned above it should be like:

INSERT INTO A_TABLE (Column_1, Column_2, Column_3,
Column_4) VALUES (Value_1,Value_2, Value_3, Column_4)

And since Column_1 is an AUTO_INCREMENT-id, u don't need to include it
in your insert at all.

So u can use this:

INSERT INTO A_TABLE (Column_2, Column_3,
Column_4) VALUES (Value_2, Value_3, Column_4)

Greets,

Bart

Nov 23 '06 #5

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