如何针对依赖用户输入的长查询在PHP中显示MySQL错误? [英] How do I display a MySQL error in PHP for a long query that depends on the user input?

问题描述

在PHP中，我试图执行一个长的MySQL查询，该查询取决于用户输入.但是，我的查询失败，并显示以下消息，

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".

实际上，每当查询失败时，我都会打印此消息，但是我很难找到此失败背后的原因.不幸的是，我找不到它，因为该错误未在网页上指定.有没有办法在网页上显示导致失败的错误消息?

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?

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));

推荐答案

使用此:

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

您还可以使用它来打印错误代码.

You can also use this to print the error code.

echo mysqli_errno($this->db_link);

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