为什么我的 ajax 不能从数据库中获取数据？[关闭]

问：

当我输入 SYMPTOMS 而不是得到 Illness 时，它会显示 else 语句

<?php
$conn = mysqli_connect("localhost" , "root" , "" , "symptoms");
if (!$conn) {
      die("Connection failed: " . mysqli_connect_error());
  }

$getMesg = mysqli_real_escape_string($conn, $_POST['text']);


$check_data = "SELECT * FROM diagnose2 WHERE Symptoms LIKE '%diagnose2%'";
$run_query = mysqli_query($conn, $check_data) or die("Error");

if(mysqli_num_rows($run_query) > 0){
    
    $fetch_data = mysqli_fetch_assoc($run_query);
    $replay = $fetch_data['Illness'];
    echo $replay;
}else{
    echo "Sorry can't be able to understand you!";
}

$conn->close();

?>

而不是 Sorry can't able to understand you！，显示 Illness