Writing 2 records on insert

[jkappl]

I have one insert statement which is supposed to produce one record in the database, but instead it produces two. My program is simple in that I open the database, do one insert, and close the database.



The two records are identical except for the auto-incremented integer which is a primary key, and one of the columns containing a unix timestamp integer value inserted as simply an integer. I obtain the value before trying INSERT and try INSERT only once (verified by before and after print values) .

It is a mystery how the constant value for the time can change between the two records while I cannot locate where the second record is being written.

Please Help ....

 

[sleipnir214]

An insert statement doesn't necessarily insert only one record. It can insert any number.

For example, in a table "fubar" with two columns "foo" and "bar", this query:

INSERT INTO fubar (foo, bar) VALUES (1, 'snafu'), (2, 'fubar')

will insert two records.

Could something like that be happening in your query?

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http://www.catb.org/~esr/faqs/smart-questions.html

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[jkappl]

Your thought is the common way that one statement might write two records. This not what is happening.

It almost seems that the SQL interpreter is playing with me. I can use the same insert statement in a different (much smaller file) and it works as it should. Same statement, cut and pasted to a different file, but now it works.

I was looking for any experience that you might have that suggests that the interpreter can do this sort of thing. And how one keeps the code clean so as to avoid this sort of anomaly.

Here is the code.


$link = mysql_connect("localhost", "peak-boo", "jude01&quot
or die("Could not connect: " . mysql_error());
// print ("Connected successfully"

mysql_select_db("peakboo_FirstTry&quot
or die("Unable to select database"
// print ("Database Selected"

/* test variables */

$custom = 'no-affil';
$owners_name = 'John Doe';
$owners_address = '21 tester street, Columbus, Ohio, 34567, status = confirmed';
$option_name1 = 'bookcode';
$option_selection1 = 'UIMC';
$option_name2 = 'customer_number';
$option_selection2 = '1061193344';
$time_xx = getdate();
$sold_time = $time_xx[0];
$activation_key = 'AAAA-BBBB-CCCC-DDDD';
$payer_email = 'jdoe@peak-books.com';
$tempxx = '1234567890123456789012345678901234567890123456789012345678901234567890123456789012345678901234567890';
for( $lci = 0; $lci < 15; $lci++){
$str_acr = $str_acr.$tempxx;
}
echo '<br>before insert','$sold_time_1 = ',$sold_time;
$query = &quot;INSERT INTO ready_for_pickup &quot;;
$query .= &quot;($option_name1,custom,sold,$option_name2,owners_name,owners_address,activation_key,email,str_acr) &quot;;
$query .= &quot;VALUES('$option_selection1','$custom','$sold_time','$option_selection2','$owners_name','$owners_address','$activation_key','$payer_email','$str_acr')&quot;;

mysql_query($query)
or die(&quot;<br>Insert Failed&quot;
// print (&quot;Entry Complete&quot;
// printf (&quot;Last inserted record has id %d\n&quot;, mysql_insert_id());

echo '<br>after insert','$sold_time_2 = ',$sold_time;


mysql_close($link);

-------------------------------------
Thanks for answering !

 

[sleipnir214]

Have you output $query and examined it?

Have you copy-and-pasted the SQL created into a MySQL app to test its operation there?

Want the best answers? Ask the best questions:

http://www.catb.org/~esr/faqs/smart-questions.html

TANSTAAFL!!