The mysqli_error() function is used to return the error in the most recent MySQL function call that failed. If there are multiple MySQL function calls, the error in the last statement is the one that is pointed out by the function.
Syntax:
mysqli_error("database_name")
Parameters: This function accepts single parameter as mentioned above and described below:
- database_name: It is the database on which operations are being performed. It is a mandatory parameter.
Program 1:
<?php $conn = mysqli_connect( "localhost", "root", "", "Persons"); // Check connection if (mysqli_connect_errno()) { echo "Database connection failed."; } // Check for error in query if (!mysqli_query($link, "SET Age=1")) { printf("Error message: %s\n", mysqli_error($conn)); } mysqli_close($conn); ?> |
Suppose the operation is being carried out on the table given below:
The output will be:
Error message: Unknown system variable 'Age'
Program 2:
<?php $conn = mysqli_connect( "localhost", "root", "", "Persons"); // Check connection if (mysqli_connect_errno()) { echo "Database connection failed."; } // Check for error in query if (!mysqli_query($link, "SET Firstname='Arkadyuti'")) { printf("Error message: %s\n", mysqli_error()); } mysqli_close($conn); ?> |
Output:
Error message: mysqli_error() expects exactly 1 parameter, 0 given
This example also demonstrates that mysqli_error() needs a database as a parameter.
