php 捕获 mysql_query 错误
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Catching mysql_query error
提问by cds5059
I am trying to stop the mysql_query error from being output onto my form. Currently if there is no location found, I receive the error
我试图阻止 mysql_query 错误输出到我的表单上。目前如果没有找到位置，我会收到错误
"Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 11"
“警告：mysql_result() [function.mysql-result]：无法跳转到 MySQL 结果索引 11 上的第 0 行”
I am trying to catch this error, and instead assign the $location variable to not found. My code for attempting this is below, what am I doing wrong?
我试图捕捉这个错误，而是将 $location 变量分配给 not found。我的尝试代码如下，我做错了什么？
Thanks!
谢谢！
$query3 = "SELECT `location` FROM location WHERE vin = '$vin' LIMIT 1";
$result3 = mysql_query($query3);
if (!mysql_result($result3,0)) {
$location = "Not found";
} else $location = mysql_result($result3,0,0);
回答by Marc B
mysql_result()generally shouldn't be used. You'd be better off with something like:
mysql_result()一般不应使用。您最好使用以下内容：
$result3 = mysql_query($query3) or die(mysql_error());
if (mysql_numrows($result3) == 0) then
$location = "not found";
} else {
$row = mysql_fetch_array($result3);
$location = $row[0];
}
Your error is caused by the fact that the query returned no rows - e.g. nothing matched. You then tried to retrieve the first field in the first row of that result set, a row which doesn't exist. Checking the number of returned rows with mysql_numrows()is safer, as that works whether the query found nothing or a bajillion rows.
您的错误是由于查询没有返回任何行 - 例如没有匹配的事实引起的。然后，您尝试检索该结果集第一行中的第一个字段，该行不存在。检查返回的行数mysql_numrows()更安全，因为无论查询未发现任何内容还是大量行，这都有效。
回答by artyom.stv
First, you can add @:
首先，您可以添加@：
if (!@mysql_result($result3,0)) {
$location = "Not found";
} else $location = mysql_result($result3,0,0);
Second, you can check mysql_num_rows(result3)before mysql_resultcall.
其次，您可以mysql_num_rows(result3)在mysql_result致电前检查。
回答by picus
You should look into how to set your error and warning levels in php ini - usually you want a s little output on prod as possible.
您应该研究如何在 php ini 中设置错误和警告级别 - 通常您希望在 prod 上输出尽可能少。
http://www.php.net/manual/en/errorfunc.configuration.php#ini.error-reporting
http://www.php.net/manual/en/errorfunc.configuration.php#ini.error-reporting
However, here, the code that would generate that error is:
但是，在这里，会产生该错误的代码是：
$result3 = mysql_query($query3);
That is the line you should be writing your if or "or die" statements around:
这就是你应该写下 if 或“or die”语句的那一行：
$result3 = mysql_query($query3)or die($location = "not found");
回答by Yahel
Mysql_query returns false if nothing is found so a simple :
如果没有找到任何东西，Mysql_query 返回 false 如此简单：
$result3 = mysql_query($query3);
if (mysql_affected_rows($result3) == 0) {
$location = "Not found";
} else $location = mysql_result($result3,0,0);
Should do it.
应该做。
回答by tcarter2005
You should look into using OOP; using a database class to handle interaction with your DB.
您应该考虑使用 OOP；使用数据库类来处理与数据库的交互。
But, basically you want to check if there are any rows, before trying to bring back the results.
但是，基本上您想在尝试返回结果之前检查是否有任何行。
Try checking with "mysql_num_rows" in your "if" statement:
尝试在“if”语句中检查“mysql_num_rows”：
if (!mysql_num_rows($result3)) {
