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I"m getting

Fatal error: Uncaught Error: Call to undefined function mysql_real_escape_string() in C:\xampp\htdocs\project\trial.php:23 Stack trace: #0 {main} thrown in C:\xampp\htdocs\project\trial.php on line 23

please help me this is my 1st project so I'm not able to solve this
help me as soon as possible

What I have tried:

Insert of Row into Database Failed - #'.mysql_errno().': '.mysql_error().'

echo '

Person\'s Information Inserted


# Prepare the SELECT Query
$selectSQL = 'SELECT * FROM `guide`';
# Execute the SELECT Query
if( !( $selectRes = mysql_query( $selectSQL ) ) ){
echo 'Retrieval of data from Database Failed - #'.mysql_errno().': '.mysql_error();

while( $row = mysql_fetch_assoc( $selectRes ) ){
echo "\n";
Name Address Line 1 Address Line 2 Email Id
No Rows Returned
Updated 7-Jan-22 9:56am
Richard Deeming 10-Jan-22 9:01am    
The fact that you're using the "escape string" functions instead of a properly parameterized query suggests that you're writing code which is vulnerable to SQL Injection[^].

PHP: SQL Injection - Manual[^]
PHP: Prepared statements and stored procedures - Manual[^]

1 solution

Whatever $result is, it's not what mysqli_num_rows is expecting. It's expecting an object of type mysqli_result, but you passed in something else.

The problem is above the code you posted, where $result is set to some value or object.
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