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How to execute multiple blocks of SQL in PHP ?
Posted by: Arthur Chan
Date: November 13, 2018 10:10PM

I need to check the validity of two identifiers that the user entered via Java and sent via https to PHP.
The variables, CID and TID, are of type INT.
PHP would only execute the first block of codes to check the first variable, in this case CID, but then drops out of the script without checking the 2nd block.

What I have done so far:
Research on the internet suggested I do this
$conn = mysql_connect('localhost','user','pwd', instance, 65536)
then create a multi-statemebnt block like this
// SQL query
// SQL query
It didn't work for me b/c I cannot embed PHP processing logic betw successive SQL queries.

How do I get PHP to continue executing after the first block of codes?

Codes that I have tried, works for 1st block, then drops out of the script:
$con = mysqli_connect("localhost", "user", "pwd", "instance");
mysqli_set_charset( $con, "utf8mb4" );
if (mysqli_connect_errno($con)) {
   die('Failed to connect to MySQL : ' .mysqli_connect_error());
$cI= (int)$_POST['CI']; 
$fT= (int)$_POST['FT'];

$testCI = "SELECT CName from Company WHERE CID='$cI'";
$result = mysqli_query($con,$testCI);
$numrows = mysqli_num_rows($result);
printf("Number of rows : ". $numrows);
if (!$numrows >0){
     die("Check CID : " .mysqli_error());
// printf("Number of rows : ". $numrows);
//while ($row=mysqli_fetch_row($result)){
//     printf("%s\n",$row[0]);


$testFT = "SELECT TName from Company WHERE TID='$fT'";
$result = mysqli_query($con,$testFT);
$numrows = mysqli_num_rows($result);
printf("Number of rows : ". $numrows);
if (!$numrows >0){
     die("Check TID : " .mysqli_error());

// TODO insert record if both CID and TID are correct


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How to execute multiple blocks of SQL in PHP ?
November 13, 2018 10:10PM

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