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lplatzAsked on October 24, 2019 at 9:01 PM
Can the script (referenced below) be modified to post form data into more than one table?
How to send submissions to your MySQL database using PHP
https://www.jotform.com/help/126-How-to-send-Submissions-to-Your-MySQL-Database-Using-PHP -
roneetReplied on October 24, 2019 at 10:02 PM
Yes, you can modify the query here, you can write join table queries to insert values in multiple tables.
$query = "SELECT * FROM `table_name` WHERE `submission_id` = '$submission_id'";
Let us know if you have further questions.
Thanks.
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lplatzReplied on October 24, 2019 at 11:36 PM
roneet provided a solution which if I were more skilled in mySQL, could probably have ran with and solved my issue. Given I'm not a skilled mySQL developer, I discovered another means for posting form data from a single form into multiple tables within my mySQL database.
Using the sample php file jotform provides for posting data to your personal mySQL database, I modified the following section of this form (Ref: https://www.jotform.com/help/126-How-to-send-Submissions-to-Your-MySQL-Database-Using-PHP).
I simply replicated the "search submission ID" section for each table I wanted to post form data into. In my case, I had a single form collecting information relevant to two tables. Hopefully the following makes sense. Perhaps any mySQL developers who may run across this can hang a little meat on this solution to whether this is a good or bad approach.
// search submission ID
$query = "SELECT * FROM `table_name1` WHERE `submission_id` = '$submission_id'";
$sqlsearch = mysql_query($query);
$resultcount = mysql_numrows($sqlsearch);
if ($resultcount > 0) {
mysql_query("UPDATE `table_name1` SET
`tbl1item1` = '$tbl1item1',
`tbl1item2` = '$tbl1item2',
`tbl1item3` = '$tbl1item3',
WHERE `submission_id` = '$submission_id'")
or die(mysql_error());
} else {
mysql_query("INSERT INTO `table_name1` (submission_id, formID, IP, tbl1item1, tbl1item2, tbl1item3)
VALUES ('$submission_id', '$formID', '$ip', '$tbl1item1', '$tbl1item2', '$tbl1item3') ")
or die(mysql_error());
}
// This section for the 2nd table
$query = "SELECT * FROM `table_name2` WHERE `submission_id` = '$submission_id'";
$sqlsearch = mysql_query($query);
$resultcount = mysql_numrows($sqlsearch);
if ($resultcount > 0) {
mysql_query("UPDATE `table_name2` SET
`tbl2item1` = '$tbl2item1',
`tbl2item1` = '$tbl2item1',
WHERE `submission_id` = '$submission_id'")
or die(mysql_error());
} else {
mysql_query("INSERT INTO `table_name2` (submission_id, tbl2item1, tbl2item1)
VALUES ('$submission_id', '$tbl2item1', '$tbl2item1') ")
or die(mysql_error());
}