如何根据用户输入的长查询在PHP中显示MySQL错误？

Question

在PHP中,我试图执行一个取决于用户输入的长MySQL查询.但是,我的查询失败,并显示以下消息:

"Query Failed".

实际上,每当查询失败时我都会打印此消息,但我很难找到导致此失败的原因.不幸的是,我找不到它,因为网页上没有指定错误.有没有办法显示导致网页失败的错误消息？

这是我的代码,

$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));

Answer 1

用这个:

mysqli_query($this->db_link, $query) or die(mysqli_error($this->db_link)); 
# mysqli_query($link,$query) returns 0 if there's an error.
# mysqli_error($link) returns a string with the last error message

您也可以使用它来打印错误代码.

echo mysqli_errno($this->db_link);

看看这里和这里

Answer 2

使用函数die():

or die(mysql_error());

Answer 3

我使用以下命令打开MySQLi的所有错误报告

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

*注意：请勿在生产环境中使用此功能。