为什么在 PHP 中将 print_r() 应用于数组时会得到“Resource id #4”？
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Why do I get "Resource id #4" when I apply print_r() to an array in PHP?
提问by Steven
Possible Duplicate:
How do i “echo” a “Resource id #6” from a MySql response in PHP?
Below is the code:
下面是代码：
$result=mysql_query("select * from choices where a_id='$taskid'")or die(mysql_error());
print_r($result);
I get "Resource id #4", any idea?
我得到“资源 ID #4”，知道吗？
After I added
我添加后
while($row=mysql_fetch_assoc($result))
{ print_r($row); }
I just got []
我刚得到 []
What's wrong?
怎么了？
回答by Kenji Kina
You are trying to print a mysql resource variableinstead of the values contained within the resource it references. You must first try to extract the values you have gotten by using a function such as mysql_fetch_assoc().
您正在尝试打印 mysql资源变量，而不是它引用的资源中包含的值。您必须首先尝试使用诸如 mysql_fetch_assoc().
You might also try mysql_fetch_array()or mysql_fetch_row(), but I find associative arrays quite nice as they allow you to access their values by the field name as in Mike's example.
您也可以尝试mysql_fetch_array()或mysql_fetch_row()，但我发现关联数组非常好，因为它们允许您按Mike 的示例中的字段名称访问它们的值。
回答by Mike B
mysql_query()does not return an array as explained in the manual. Use mysql_fetch_array(), mysql_fetch_assoc(), or mysql_fetch_row()with your $result. See the link above for more info on how to manipulate query results.
mysql_query()不返回手册中解释的数组。使用mysql_fetch_array(), mysql_fetch_assoc(), 或mysql_fetch_row()与您的$result. 有关如何操作查询结果的更多信息，请参阅上面的链接。
$result = mysql_query('SELECT * FROM table');
while ($row = mysql_fetch_assoc($result)) {
echo $row["userid"];
echo $row["fullname"];
echo $row["userstatus"];
}
回答by mauris
$resultis a resource variable returned by mysql_query. More about resource variables: http://php.net/manual/en/language.types.resource.php
$result是由 返回的资源变量mysql_query。有关资源变量的更多信息：http: //php.net/manual/en/language.types.resource.php
You must use other functions such as mysql_fetch_array()or mysql_fetch_assoc()to get the array of the query resultset.
您必须使用其他函数，例如mysql_fetch_array()或mysql_fetch_assoc()来获取查询结果集的数组。
$resultset = array();
$result=mysql_query("select * from choices where a_id='$taskid'") or die(mysql_error());
while($row = mysql_fetch_assoc($result)){
$resultset[] = $row; // fetch each row...
}
mysql_free_result($result); // optional though...
print_r($resultset);
See:
看：
http://php.net/manual/en/function.mysql-fetch-array.php
http://php.net/manual/en/function.mysql-fetch-assoc.php
http://php.net/manual/en/function.mysql-query.php
http://php.net/manual/en/function.mysql-fetch-array.php
http://php.net/manual/en/function.mysql-fetch-assoc.php
http://php.net/手册/en/function.mysql-query.php
