用数组PHP制作对象

Question

我试图在整个桌子上搜索一个单词。 因此，如果您搜索Eminem，则必须使用Eminem这个单词来获取所有内容。

我搜索

<?php

 $sql="SELECT * FROM album WHERE albumartiest like '$zoek'";
    $resultaatcolumn = Yii::app()->db->CreateCommand($sql)->queryAll();                         
    if($resultaatcolumn != null){
     $zoekresultaat[] = $resultaatcolumn;}
 $sql="select * from album where albumnaam like '%$zoek%'";
    $resultaatcolumn = Yii::app()->db->CreateCommand($sql)->queryAll();
    if($resultaatcolumn != null){
     $zoekresultaat[] = $resultaatcolumn;}
 $sql="select * from album where albumartiest like '%$zoek%'";
    $resultaatcolumn = Yii::app()->db->CreateCommand($sql)->queryAll();
    if($resultaatcolumn != null){
     $zoekresultaat[] = $resultaatcolumn;}
 $sql="select * from album where albumgenre like '%$zoek%'";
    $resultaatcolumn = Yii::app()->db->CreateCommand($sql)->queryAll();
    if($resultaatcolumn != null){
     $zoekresultaat[] = $resultaatcolumn;}
 $sql="select * from album where albumafspeelijst like '%$zoek%'";
    $resultaatcolumn = Yii::app()->db->CreateCommand($sql)->queryAll();
    if($resultaatcolumn != null){
     $zoekresultaat[] = $resultaatcolumn;}

它可以工作，但不完全是我想要的。 结果是这样的：

Array ( [0] => Array ( [0] => Array ( [albumcode] => 45 [albumnaam] => recovery [albumafspeelijst] => ["Cold Wind Blows","Talkin' 2 Myself","On Fire","Won't Back Down","W.T.P.","Going Through Changes","Not Afraid","Seduction","No Love","Space Bound","Cinderella Man","To Life","So Bad","Almost Famous","Love The Way You Lie","You're Never Over",""] [albumartiest] => Eminem [albumgenre] => hip-hop [albumimage] => images\eminemrecovery.png [albumprijs] => 20 ) ) [1] => Array ( [0] => Array ( [albumcode] => 45 [albumnaam] => recovery [albumafspeelijst] => ["Cold Wind Blows","Talkin' 2 Myself","On Fire","Won't Back Down","W.T.P.","Going Through Changes","Not Afraid","Seduction","No Love","Space Bound","Cinderella Man","To Life","So Bad","Almost Famous","Love The Way You Lie","You're Never Over",""] [albumartiest] => Eminem [albumgenre] => hip-hop [albumimage] => images\eminemrecovery.png [albumprijs] => 20 ) ) )

没关系，但是我想要的是取出变量并使用它。

有没有一种方法可以让我从数组中获取变量并使用它？ 如果你们想要有关我的代码的更多信息，请询问！

Answer 1

试试这个

Yii::app()->db->CreateCommand($sql)->setFetchMode(PDO::FETCH_OBJ)->queryAll()

这将为您提供一个以列名作为属性的对象数组。

例如：-

foreach($result as $row)
{
echo $row->albumcode;
}

Answer 2

如果要像对象一样访问结果集，则可以使用本机PHP类ArrayObject并提供标志来表明这一点。

$album = new ArrayObject($result, ArrayObject::ARRAY_AS_PROPS);

现在，您可以像以下那样访问结果：

$code = $album->albumcode;
$name = $album->albumnaam;

希望这可以指导您，编码愉快！

Answer 3

嗯，只是做

foreach($zoekresultaat as $key => $value) {
   //do what I want with each seperate returened result. The array key is in $key and the result array is in $value
   echo $value['albumcode'] . ' = '. $value['albumnaam'];

}

aka，基本的PHP

并请确保您应用的安全性，以了解如何在yii中执行准备好的语句

现在您查询的方式可以清除整个数据库