PHP：检查数组是否全部为真

debugcn 发表于 Dev

韦伯先生

我正在尝试缩短代码，我认为使用数组是最好的方法。

这段代码几乎是由许多不同的查询（mysqli）组成的，如下所示：

$a = $this->db->query("DELETE * FROM ....");
$b = $this->db->query("DELETE * FROM ....");
$c = $this->db->query("DELETE * FROM ....");
$d = $this->db->query("DELETE * FROM ....");
$e = $this->db->query("DELETE * FROM ....");
$f = $this->db->query("DELETE * FROM ....");
$g = $this->db->query("DELETE * FROM ....");
$h = $this->db->query("DELETE * FROM ....");
$i = $this->db->query("DELETE * FROM ....");
$j = $this->db->query("DELETE * FROM ....");
$k = $this->db->query("DELETE * FROM ....");

要检查此查询是否正确执行，我这样做（因为它们将返回true / false）：

if($a && $b && $c && $d && $e && $f && $g && $h && $i && $j && $k){
    //action    
}

但是每次我必须添加查询时，我都必须将其添加到if语句中，并且它变得很长，我不知道是否包含了它。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。

因此，现在我正在考虑使用数组执行此操作：

$a[0] = $this->db->query("DELETE * FROM ....");
$a[1] = $this->db->query("DELETE * FROM ....");
$a[2] = $this->db->query("DELETE * FROM ....");
$a[3] = $this->db->query("DELETE * FROM ....");
$a[4] = $this->db->query("DELETE * FROM ....");
$a[5] = $this->db->query("DELETE * FROM ....");
$a[6] = $this->db->query("DELETE * FROM ....");
$a[7] = $this->db->query("DELETE * FROM ....");
$a[8] = $this->db->query("DELETE * FROM ....");
$a[9] = $this->db->query("DELETE * FROM ....");
$a[10] = $this->db->query("DELETE * FROM ....");

if($a === true){
    //ok
}else{
    //notok;
}

但是此代码（如果语句）不起作用...我在做什么错？

克劳

尝试：

$queries=array(
  'DELETE * FROM ....',
  'DELETE * FROM ....',
  'DELETE * FROM ....',
  'DELETE * FROM ....',
);

$is_all_good=true;
foreach($queries as $sql) 
  if(!$this->db->query($sql)) {$is_all_good=false;break;}

echo $is_all_good?'They are all ok.':'They are not all ok.';

如果您的设置允许交易，也可以将此想法与交易合并。

$queries=array(
  'DELETE * FROM ....',
  'DELETE * FROM ....',
  'DELETE * FROM ....',
  'DELETE * FROM ....',
);

$this->db->autocommit(false);

foreach($queries as $sql) $this->db->query($sql); 

echo $this->db->commit()?'They are all ok.':'They are not all ok.';

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