如何使用mysqli获得相同的结果? [英] How can this be done with the same result using mysqli?

问题描述

我正在尝试将一些代码从mysql_移植到mysqli_等效项，但是我偶然发现了一些似乎不是直接转换的内容:

I am working on porting some code over from mysql_ to mysqli_ equivalents, however I stumbled accross something that doesn't seem to be a straight conversion:

我有以下代码

if($field_flags == false) {
    for($j=0; $j<$num_fields; $j++) {
        $field_flags[$j] = mysql_field_flags($result, $j);
        if(strpos($field_flags[$j], 'not_null') !== false) {
            $field_flags[$j] = false;
        } else {
            $field_flags[$j] = true;
        }
    }
}

由于新方法"mysqli_fetch_field_direct"似乎不太简单，我无法移植它.

And I am having trouble porting it as the new method "mysqli_fetch_field_direct" doesn't seem to be as straightforward.

有人可以协助我移植上面的代码.本质上，我正在检查该字段是否允许空值.

Could someone please assist me with porting the above code. Essentially, I am checking if the field allows nulls or not.

推荐答案

函数mysqli_fetch_field_direct返回object，而mysql_field_flags则返回string.因此，直接使用strpos无效.

The function mysqli_fetch_field_direct returns an object, where as mysql_field_flags retuns a string. Hence your direct use of strpos won't work.

if($field_flags == false) {
    for($j=0; $j<$num_fields; $j++) {
        $field_flags[$j] = mysqli_fetch_field_direct($result, $j);
        if(strpos($field_flags[$j]->flags, 'NOT_NULL') !== false) {
           //Do something here
        } else {
           //Do something else
    }
}

更多信息，请参见 php.net

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