一尘不染

php mysqli准备的语句喜欢

php

我如何使用mysqli用LIKE进行查询并获得所有结果?

这是我的代码,但是不起作用:

$param = "%{$_POST['user']}%";
$stmt = $db->prepare("SELECT id,Username FROM users WHERE Username LIKE ?");
$stmt->bind_param("s", $param);
$stmt->execute();
$stmt->bind_result($id,$username);
$stmt->fetch();

此代码似乎不起作用。我已经搜索了很多。也可能返回多于1行。那么,即使返回的结果超过1行,我如何获得所有结果呢?


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2020-05-26

共1个答案

一尘不染

这是您正确获取结果的方式

$param = "%{$_POST['user']}%";
$stmt = $db->prepare("SELECT id,Username FROM users WHERE Username LIKE ?");
$stmt->bind_param("s", $param);
$stmt->execute();
$stmt->bind_result($id,$username);

while ($stmt->fetch()) {
  echo "Id: {$id}, Username: {$username}";
}

或者您也可以:

$param = "%{$_POST['user']}%";
$stmt = $db->prepare("SELECT id,Username FROM users WHERE Username LIKE ?");
$stmt->bind_param("s", $param);
$stmt->execute();

$result = $stmt->get_result();
while ($row = $result->fetch_array(MYSQLI_NUM)) {
  foreach ($row as $r) {
    print "$r ";
  }
  print "\n";
}
2020-05-26