我确信,这并不是ug笑,而是在我利用我的智慧来挽救成果的时候,我这样做:
$result->fetch_array();
mysqli_result Object (
[current_field] => 0
[field_count] => 1
[lengths] => Array (
[0] => 3
)
[num_rows] => 1
[type] => 0
) Fatal error: Cannot use object of type mysqli_result as array
在查阅这些文件时,它要么归还一个阵列,要么放弃,而不是反对。 这难道是一种ug吗?
在研究这一法典时,应注意:
$query="SELECT * FROM table where id= ";
$result=$link->query($query);
$row=$result->fetch_array();
print_r($row);
sample代码没有在PHP 5.3.4中复制这一问题(所有样本代码都应照做)。
$result。 请使用<代码>$row。
您必须把<代码>(单位:美元-和>;fetch_array()的回报价值分配给一个后来变成一个阵列或无效的变量。
e.g.
$row = $result->fetch_array();
$row[0] etc etc.
http://www.php.net/manual/en/class.mysqli-result.php” rel=“nofollow”>mysqli 结果:
我的SQLi结果是:
系从查询数据库中获得的结果
在使用PHP阅读MySQL时,即使用:
$query = mysql_query("select something from something");
while($row = mysql_fetch_array($query)){
$msql[$i]=$row;
$i++;
}
并且现在,有两维面的薪-包,有选定的数值。
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