[SOLVED] mySQL query INSERT not working, and I'm going to go crazy!

[phpfreaksftsolitude]

====================

MY CODE.  I'm trying to insert a record into a table in my database.  It doesn't really get any simpler than this!  But for some reason it's not working no matter what I try, I get the error message below.

====================

 

$sql = "INSERT INTO phonebook (name) VALUES ('Dave')";

 

if (mysql_query($sql)) {

echo("Your record has been added.");

} else {

echo("Error adding submitted record: " . mysql_error());

}

 

 

====================

ERROR MESSAGE (below).  I feel like I've checked the syntax a hundred times.  Can somebody help me?  Thanks!

====================

 

Error adding submitted record: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'character (name) VALUES ('Dave')' at line 1

 

 

[trq]

Put simply... there is nothing wrong with that code.

[mmarif4u]

Try this:

$sql = "INSERT INTO phonebook (name) VALUES ('Dave')";
$result=mysql_query($sql);
if ($result) {
echo("Your record has been added.");
} else {
echo("Error adding submitted record: " . mysql_error());
}

[tauchai83]

I din see anything wrong. But you may try the one posted by mmarif4u 1st.

 

 

[trq]

Try this:

$sql = "INSERT INTO phonebook (name) VALUES ('Dave')";
$result=mysql_query($sql);
if ($result) {
echo("Your record has been added.");
} else {
echo("Error adding submitted record: " . mysql_error());
}

 

That is exactly the same.

[phpfreaksftsolitude]

Thanks guys, but none of those worked.  Could it have anything to do with the fact that I'm using Yahoo?  Or maybe the auto increment for the "id" field isn't working like it should?  I dunno, I'm am totally out of ideas....

 

 

[phpfreaksftsolitude]

I'm close to figuring it out!!!  Okay, it DOES have something to do with my database table.  I created a new empty table, didn't do anything fancy to it, and the insert function worked fine!  (I can't believe I spent HOURS on this, dang).

 

So I think the problem with my existing table is in the "auto increment" feature.  In the table, for the "id" row, I have set the following:

 

Type = int

Attributes = UNSIGNED

Extra = auto_increment

 

What else do I need to make this auto increment feature work correctly?  Thanks in advance....

 

 

[StarBuck]

normally INSERT queries are like "INSERT INTO table VALUES('','bob','smith')"

 

double single quotes would be the ID you want to increment and bob would be first name field, smith last name field. Thats old school though you might be using a newer method.

[SkyRanger]

Yeah, I had the same problem with it inserting and getting the same errors.  best thing to do is do the "insert into table (id,name) VALUES ('', '$name')"

 

If that doesn't work, do a manual insert through phpMyAdmin and check the code that is used and just apply it to your script.

[neel_basu]

in auto_increment ed fields you doesn't need to enter any values it will get incremented automatically.

Try this

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

And make sure is the auto_incremented field Your PRIMARY KEY ??? ??? ??

[phpfreaksftsolitude]

It is set as my primary key (I am using phpMyAdmin to set up the database).  But still no luck.  It's strange, I can insert into to the new table (which has no primary key yet) but I still cannot insert into the existing table.

 

Should I just delete the "id" field and add a new one to the table?  Should I make another field my primary?

 

 

[neel_basu]

You tried

$result=mysql_query($sql) or die(mysql_error()); ??

[phpfreaksftsolitude]

Yes, I tried that.  That's my whole problem, it works in for one table but not the other.  Why would the same code work to insert a record into one table and not in another within the same database?

[tauchai83]

could you please post your DB design? all the fields...and possibly post the whole code doesn't work.

[phpfreaksftsolitude]

I am going to close this issue because I ended up deleting the table and all its records (like I said, it was driving me crazy).  So I recreated a new table (more carefully this time) and tested it along the way, and it works perfectly!  No idea why the first table caused me so much pain.  Now I've just gotta enter in my old records one by one, which isn't too big of a deal.

 

Thanks for all your help, you guys helped me flesh out my query structure and that's what counts!

 

-S