您也尝试过concat建议（以及没有%？？
我也测试了一遍

<?php
$conn = new mysqli('localhost','root','','new_schema');
$stmt = $conn->stmt_init();
$query = "select * from csv where rgion = ? and city like concat('%',?,'%')";
$stmt->prepare($query);
$reg = 11;
$ci = 'dg';
$stmt->bind_param('is',$reg,$ci);
$stmt->bind_result($country,$city,$accentcity,$region,$population,$latitude,$longitude);
$stmt->execute();
echo "<table border='1'>";
while ($stmt->fetch()) {
    echo "<tr>";
    echo "<td>$country</td>";
    echo "<td>$city</td>";
    echo "<td>$accentcity</td>";
    echo "<td>$region</td>";
    echo "<td>$population</td>";
    echo "<td>$latitude</td>";
    echo "<td>$longitude</td>";
    echo "</tr>";
}
echo "</table>";

首先，您在bind_param中有错误的类型序列（第一位是int，第二位是string），所以请尊重类型序列

$query = "select * from csv where rgion = ? and city like ?)";
  $stmt->prepare($query);
  $reg = 11;
  $ci = '%dg%';
  $stmt->bind_param('is',$reg, $ci);

作为一个建议，试着用这种方式避免字符串中的wildchar

$query = "select * from csv where rgion = ? and city like concat('%',?,'%')";
  $stmt->prepare($query);
  $reg = 11;
  $ci = 'dg';
  $stmt->bind_param('is',$reg, $ci);

我又改了代码没用！！警告：mysqli_stmt:：bind_param（）：对象或资源mysqli_stmt无效

<?php
$conn = new mysqli('localhost','root','','new_schema');
$stmt = $conn->stmt_init();
$query = "select * from csv where rgion = ? and city like ?";
$stmt->prepare($query);
$reg = 11;
$ci = '%dg%';
$stmt->bind_param('is',$reg,$ci);
$stmt->bind_result($country,$city,$accentcity,$region,$population,$latitude,$longitude);
$stmt->execute();
echo "<table border='1'>";
while ($stmt->fetch()) {
    echo "<tr>";
    echo "<td>$country</td>";
    echo "<td>$city</td>";
    echo "<td>$accentcity</td>";
    echo "<td>$region</td>";
    echo "<td>$population</td>";
    echo "<td>$latitude</td>";
    echo "<td>$longitude</td>";
    echo "</tr>";
}
echo "</table>";