如何在 PHP 中为依赖于用户输入的长查询显示 MySQL 错误? [英] How do I display a MySQL error in PHP for a long query that depends on the user input?
本文介绍了如何在 PHP 中为依赖于用户输入的长查询显示 MySQL 错误?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
在 PHP 中,我试图执行一个依赖于用户输入的长 MySQL 查询.但是,我的查询失败并显示以下消息,
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,但我很难找到此失败背后的原因.不幸的是,我找不到它,因为网页上没有指定错误.有没有办法在网页上显示导致失败的错误信息?
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($link) returns a string with the last error message
您也可以使用它来打印错误代码.
You can also use this to print the error code.
echo mysqli_errno($this->db_link);
这篇关于如何在 PHP 中为依赖于用户输入的长查询显示 MySQL 错误?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!
查看全文