获取带有 mysqli 结果的行数组
我需要从结果对象中获取所有行.我正在尝试构建一个包含所有行的新数组.
I need to get all the rows from result object. I’m trying to build a new array that will hold all rows.
这是我的代码:
$sql = new mysqli($config['host'],$config['user'],$config['pass'],$config['db_name']);
if (mysqli_connect_errno())
{
printf("Connect failed: %s
", mysqli_connect_error());
exit();
}
$query = "SELECT domain FROM services";
$result = $sql->query($query);
while($row = $result->fetch_row());
{
$rows[]=$row;
}
$result->close();
$sql->close();
return $rows;
$rows
应该是包含所有行的新数组,但我得到了一个空数组.
$rows
is supposed to be the new array that contains all, rows but instead I get an empty array.
知道为什么会这样吗?
推荐答案
你有一个轻微的语法问题,即一个错误的分号.
You had a slight syntax problem, namely an errant semi-colon.
while($row = $result->fetch_row());
注意到末尾的分号了吗?这意味着后面的块没有在循环中执行.摆脱它,它应该可以工作.
Notice the semi-colon at the end? It means the block following wasn't executed in a loop. Get rid of that and it should work.
另外,你可能想让mysqli报告它遇到的所有问题:
Also, you may want to ask mysqli to report all problems it encountered:
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$sql = new mysqli($config['host'], $config['user'], $config['pass'], $config['db_name']);
$query = "SELECT domain FROM services";
$result = $sql->query($query);
$rows = [];
while($row = $result->fetch_row()) {
$rows[] = $row;
}
return $rows;
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