Can anyone here help me on how can I use PHP
to insert multiple rows into a database with a single submit? I have tried doing it, but it only inserts one row.
Here is my code:
<?php
if(isset($_POST['insertData']))
{
$pred1 =$_POST['pre'];
$np1 =$_POST['nap'];
$sd101 =$_POST['tdisease'];
$pr1 =$_POST['pric1'];
$ivd =$_POST['invd'];
$id =$_POST['user'];
$pred1 =$_POST['pre1'];
$np1 =$_POST['nap1'];
$sd101 =$_POST['tdisease1'];
$pr1 =$_POST['pric1'];
$pred2 =$_POST['pre2'];
$np2 =$_POST['nap2'];
$sd102 =$_POST['tdisease2'];
$pr2 =$_POST['pric2'];
$insert_user="INSERT INTO invoices(id, icd10, nappi_code, prescription, price, invoice_date, pid) VALUES ('','$sd10', '$np' ,'$pred','$pr','$ivd','$id');";
$insert_user .="INSERT INTO invoices(id, icd10, nappi_code, prescription, price, invoice_date, pid) VALUES ('','$sd101', '$np1' ,'$pred1','$pr1','$ivd','$id');";
$insert_user .="INSERT INTO invoices(id, icd10, nappi_code, prescription, price, invoice_date, pid) VALUES ('','$sd102', '$np2' ,'$pred2','$pr2','$ivd','$id')";
if(mysqli_query($con,$insert_user))
{
echo"<script>alert(' Invoice Details successfuly added to database')</script>";
echo '<meta content="1;generate-invoive-results-date-report.php?id='.$id.'" http-equiv="refresh" />';// redirects user view page after 3
}else{
echo"<script>alert('Unknown error occured')</script>";
}
}
?>
My guess is that the first insert completes, and then the subsequent two are being ignored. But why don't you insert all data in a single statement?
$insert_user="INSERT INTO invoices(id, icd10, nappi_code, prescription, price, invoice_date, pid) VALUES ('','$sd10', '$np' ,'$pred','$pr','$ivd','$id'),";
$insert_user .="('','$sd101', '$np1','$pred1','$pr1','$ivd','$id'),";
$insert_user .="('','$sd102', '$np2' ,'$pred2','$pr2','$ivd','$id');";
You should try this perhaps:
$insert_user="INSERT INTO invoices(id, icd10, nappi_code, prescription, price, invoice_date, pid) VALUES ('','$sd10', '$np' ,'$pred','$pr','$ivd','$id'),('','$sd101', '$np1' ,'$pred1','$pr1','$ivd','$id'), ('','$sd102', '$np2' ,'$pred2','$pr2','$ivd','$id')";
And also, This:
$pred1 =$_POST['pre'];
$np1 =$_POST['nap'];
$sd101 =$_POST['tdisease'];
$pr1 =$_POST['pric1'];
$ivd =$_POST['invd'];
$id =$_POST['user'];
$pred1 =$_POST['pre1'];
$np1 =$_POST['nap1'];
$sd101 =$_POST['tdisease1'];
$pr1 =$_POST['pric1'];
$pred2 =$_POST['pre2'];
$np2 =$_POST['nap2'];
$sd102 =$_POST['tdisease2'];
$pr2 =$_POST['pric2'];
Should've been:
$ivd =$_POST['invd'];
$id =$_POST['user'];
$pred =$_POST['pre'];
$np =$_POST['nap'];
$sd10 =$_POST['tdisease']; <==
$pr =$_POST['pric1'];
$pred1 =$_POST['pre1'];
$np1 =$_POST['nap1'];
$sd101 =$_POST['tdisease1'];
$pr1 =$_POST['pric1'];
$pred2 =$_POST['pre2'];
$np2 =$_POST['nap2'];
$sd102 =$_POST['tdisease2'];
$pr2 =$_POST['pric2'];
Because the first two sets have the same variable names and they'll be overwritten.
====================== Edit ===========================
The reason why it didn't work for you before was probably because:
1) The variables '$sd10', '$np' ,'$pred','$pr'
didn't exist in the 1st query:
$insert_user="INSERT INTO invoices(id, icd10, nappi_code, prescription, price, invoice_date, pid) VALUES ('','$sd10', '$np' ,'$pred','$pr','$ivd','$id');";
2) There was no ;
at the end of the 3rd query:
VALUES ('','$sd102', '$np2' ,'$pred2','$pr2','$ivd','$id')<no-semi-colon-here>";
As they are being inserted as 3 separate queries.
A query need not be given all on a single line, so lengthy queries that require several lines are not a problem. mysql determines where your statement ends by looking for the terminating semicolon, not by looking for the end of the input line. (In other words, mysql accepts free-format input: it collects input lines but does not execute them until it sees the semicolon.)
Refer: https://dev.mysql.com/doc/refman/5.7/en/entering-queries.html
You should have a look at the function mysqli_multi_query . See example here : https://www.w3schools.com/php/php_mysql_insert_multiple.asp
Hope this helps
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