[php] How do I display a MySQL error in PHP for a long query that depends on the user input?

In PHP, I am trying to execute a long MySQL query that depends on the user input. However, my query fails with the following message,

"Query Failed".

Actually I have printed this message whenever the query fails, but I am having hard time looking for the reason behind this failure. Unfortunately, I couldn't find it because the error is not specified on the web page. Is there a way to display the error message that caused the failure on the web page?

Here's my code,

$from = "Findings";
$where = "";

if ($service != null)
{
    $from = $from . ", ServiceType_Lookup";
    $where= "Findings.ServiceType_ID= ServiceType_Lookup.ServiceType_ID AND ServiceType_Name= ". $service;

    if ($keyword != null)
        $where= $where . " AND ";
}

if ($keyword != null)
{
    $where= $where . "Finding_ID LIKE '%$keyword%' OR
                     ServiceType_ID LIKE '%$keyword%' OR
                     Title LIKE '%$keyword%' OR
                     RootCause_ID LIKE '%$keyword%' OR
                     RiskRating_ID LIKE '%$keyword%' OR
                     Impact_ID LIKE '%$keyword%' OR
                     Efforts_ID LIKE '%$keyword%' OR
                     Likelihood_ID LIKE '%$keyword%' OR
                     Finding LIKE '%$keyword%' OR
                     Implication LIKE '%$keyword%' OR
                     Recommendation LIKE '%$keyword%' OR
                     Report_ID LIKE '%$keyword%'";
}

$query = "SELECT Finding_ID,
                 ServiceType_ID,
                 Title,
                 RootCause_ID,
                 RiskRating_ID,
                 Impact_ID,
                 Efforts_ID,
                 Likelihood_ID,
                 Finding,
                 Implication,
                 Recommendation,
                 Report_ID  FROM ".$from . " WHERE " . $where;

echo "wala 2eshiq";

$this->result = $this->db_link->query($query);
if (!$this->result) {
    printf("Query failed: %s\n", mysqli_connect_error());
    exit;
}

$r = mysqli_query($this->db_link, $query);
if ($r == false)
    printf("error: %s\n", mysqli_errno($this->db_link));

Use below code to print the error code :

echo mysqli_errno($this->db_link);

Error code will give you better idea about the error.

More info can be found at https://www.techqura.com/techqura.php?post=How-to-display-MySQL-error-in-PHP&pid=8&website=techqura.com

Try something like this:

$link = @new mysqli($this->host, $this->user, $this->pass)
$statement = $link->prepare($sqlStatement);
                if(!$statement)
                {
                    $this->debug_mode('query', 'error', '#Query Failed<br/>' . $link->error);
                    return false;
                }

I use the following to turn all error reporting on for MySQLi

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

*NOTE: don't use this in a production environment.

The suggestions don't work because they are for the standard MySQL driver, not for mysqli:

$this->db_link->error contains the error if one did occur

Or

mysqli_error($this->db_link)

will work.

Use function die():

or die(mysql_error());

One useful line of code for you would be:

$sql = "Your SQL statement here";
$result = mysqli_query($this->db_link, $sql) or trigger_error("Query Failed! SQL: $sql - Error: ".mysqli_error($this->db_link), E_USER_ERROR);

This method is better than die, because you can use it for development AND production. It's the permanent solution.