如何在 PHP 中为依赖于用户输入的长查询显示 MySQL 错误？
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How do I display a MySQL error in PHP for a long query that depends on the user input?
提问by Traveling Salesman
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,
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，
"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?
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？
Here's my code,
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($this->db_link));
回答by Christian
Use this:
用这个：
mysqli_query($this->db_link, $query) or die(mysqli_error($this->db_link));
# mysqli_query($link,$query) returns 0 if there's an error.
# mysqli_error($link) returns a string with the last error message
You can also use this to print the error code.
您也可以使用它来打印错误代码。
echo mysqli_errno($this->db_link);
回答by Arian Faurtosh
I use the following to turn all error reporting on for MySQLi
我使用以下命令为 MySQLi 打开所有错误报告
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
*NOTE:don't use this in a production environment.
*注意：不要在生产环境中使用它。
回答by mc_fish
The suggestions don't work because they are for the standard MySQL driver, not for mysqli:
这些建议不起作用，因为它们适用于标准 MySQL 驱动程序，而不适用于 mysqli：
$this->db_link->errorcontains the error if one did occur
$this->db_link->error如果确实发生了错误，则包含错误
Or
或者
mysqli_error($this->db_link)
will work.
将工作。
回答by Abdel Rahman Karim
Try something like this:
尝试这样的事情：
$link = @new mysqli($this->host, $this->user, $this->pass)
$statement = $link->prepare($sqlStatement);
if(!$statement)
{
$this->debug_mode('query', 'error', '#Query Failed<br/>' . $link->error);
return false;
}
回答by Hrvoje Antunovi?
One useful line of code for you would be:
对您有用的一行代码是：
$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.
此方法优于die，因为您可以将其用于开发和生产。这是永久解决方案。
