捕捉Mysqli错误 [英] Catching Mysqli Errors

问题描述

我有一个自定义的错误处理程序定义，它捕获所有异常并将它们保存到一个日志中。现在，如果我在mysqli查询中有语法错误，比如打字错误，那么该页面就完全停止加载。抛出异常，因此错误处理程序不被触发，没有任何记录。

这是否正常？有没有PHP设置，我应该检查来修复这个，以便任何mysqli查询错误引发异常？

（如果查询返回0，我不希望抛出异常结果 - 只有由于查询结构中的打字错误或其他错误而导致错误）

示例查询：

< $（$ r = $ result-> fetch_assoc（））{$ data =$ ] = $ R;}} $ result->免费（）;

如果我尝试在PHPMyAdmin中执行此查询，它会告诉我列bad_field_reference不存在。如果我尝试执行它作为我的PHP脚本的一部分，整个页面停止加载在那一点。

澄清我只是在做通过查看页面源码进行一些测试。显然，页面的其余部分加载 - 但是，当启用java时，某些项目被隐藏，那么我使用jquery以动画方式重新显示一些内容。这些jquery脚本没有运行，所以该页面似乎已经停止加载。

所以 - 现在有2个问题 - 如何让PHP抓住错误，如何让jquery仍然运行它的脚本？

解决方案

这是正常吗？

否。

有没有PHP设置我应该检查以修复这个，以便任何mysqli查询错误抛出异常？

是。

$ b $ （$ MYSQLI_REPORT_STRICT）

但是，并不是所有的mysqli_ *函数都会引发DB错误，其中一些引发了常规的PHP错误，像 free（）这样做，这部分让我想起更多：我也很好奇，为什么只处理例外，但是没有处理PHP错误。不要告诉我你有error_reporting（0）; - 你吗？如果是这样 - 我没有话。

最后，我怀疑你有原始的mysqli-> query（）分散在整个代码中。这意味着

• 您没有占位符来构建查询 - 使SQL注入不可避免。


• 你的代码blo肿和不可读

你必须使用一个抽象库将所有重复的代码转换为类方法。使用 safeMysql 查看运行的查询：

  $ data = $ db-> getAll（'SELECT bad_field_reference FROM table'）; 
  

它不需要水平滚动来阅读它，它是安全，简洁，防错误和支持占位符！

I have a custom error handler defined, which catches all exceptions and saves them to a log. Right now, if I have a syntax error in a mysqli query, such as a typo, the page completely stops loading at that point. No exception is thrown, so therefore the error handler isn't triggered and nothing is logged.

Is this normal? Is there a PHP setting that I should check to fix this so that any mysqli query errors throw exceptions?

(I don't want an exception thrown if a query returns 0 results - only if it errors out due to a typo or other error in the query structure)

Example query:

if($result=$db->query('SELECT bad_field_reference FROM table')){while($r=$result->fetch_assoc()){$data[]=$r;}}$result->free();

If I try to execute this query in PHPMyAdmin it will tell me that column bad_field_reference does not exist. If I try to execute it as part of my PHP script, the entire page stops loading at that point.

To clarify I was just doing some testing by looking at the page source. Apparently the rest of the page does load - however, when java is enabled certain items are hidden, then I use jquery to redisplay some content in an animated fashion. Those jquery scripts are not running, so the page appears to have stopped loading.

So - 2 questions now - how do I get PHP to 'catch' the error, and how do I get jquery to still run it's scripts?

解决方案

Is this normal?

No.

Is there a PHP setting that I should check to fix this so that any mysqli query errors throw exceptions?

Yes.

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

However, not all the mysqli_* functions would raise a DB error, some of them raise regular PHP errors, like free() does, and this part makes me wonder even more: I am also extremely curious, why do you handle exceptions only but leave PHP errors unhandled. Don't tell me you have error_reporting(0); - do you? If so - I've no words.

Finally, I have suspicion that you have raw mysqli->query() scattered all over the code. Which means

• you are using no placeholders to build your queries - making SQL injection inevitable.
• your code is bloated and unreadable

You have to use an abstraction library that will put all the repeated code into class methods. See your query run with safeMysql:

$data = $db->getAll('SELECT bad_field_reference FROM table');

it require no horizontal scrolling to read it, it is safe, concise, error-proof and supports placeholders!

这篇关于捕捉Mysqli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！