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Php mysql对象未转换为int_Php_Mysql - Fatal编程技术网

Php mysql对象未转换为int

Php mysql对象未转换为int,php,mysql,Php,Mysql,因此,我有一个wp mysql db列,如下所示: `update_number` = int(11) Null:No, default:0 然后我有以下php: $RSS_UPDATE = $wpdb->get_results("SELECT update_number FROM $table WHERE sub_id = $post_id"); if ( $RSS_UPDATE ) { foreach ( $RSS_UPDATE as $RSS_SINGLE ) {

因此,我有一个wp mysql db列,如下所示:

`update_number` =  int(11) Null:No, default:0
然后我有以下php:

$RSS_UPDATE = $wpdb->get_results("SELECT update_number FROM $table WHERE sub_id = $post_id");
if ( $RSS_UPDATE )
{
    foreach ( $RSS_UPDATE as $RSS_SINGLE )
    {
        $RSS_ROW_NEW  = $RSS_SINGLE + 1;            
        $wpdb->update($table, array('update_number' => $RSS_ROW_NEW),array( 'sub_id' => $post_id ));    
    }   
}
因此,我们的想法是,对于
sub\u id=$post\u id
的任何结果,然后获取所有
update\u编号
。(即可以是“0,2,1,4,2,2”)

然后,对于每个值,我想将整数值增加+1并更新它

但是,我得到的stdClass类的
对象无法转换为int


我做错了什么?

我认为您需要确保$RSS\u SINGLE是string还是int

var_dump($RSS_SINGLE);退出

$RSS\u UPDATE=$wpdb->获取结果(“从$table中选择UPDATE\u number,其中sub\u id=$post\u id”);
如果($RSS\u更新)
{
foreach($RSS\u更新为$RSS\u单)
{
$RSS\u ROW\u NEW=$RSS\u SINGLE->update\u number+1;
$wpdb->update($table,array('update\u number'=>$RSS\u ROW\u NEW),array('sub\u id'=>$post\u id));
}   

}
在循环内部执行变量转储($RSS\u SINGLE);退出看看你得到了什么。这可能是个愚蠢的问题,但我如何正确使用
var\u dump($RSS\u SINGLE);退出?我在哪里看到结果?有没有办法在
debug
文件中显示?在$RSS\u ROW\u NEW=$RSS\u SINGLE->update\u number+1之前;