未答复的PHP - 选择树状况列并显示其他3列

Question

I want to display treeid, treelatitude and treelongitude by selecting the treecondition column.

When I POST "healthy" I am able to display treeid 4,6 & 7, But If I POST "Balanced" Not able to get treeid 1 & 8. How can I get it?

In my below code I can only able to display "Healthy" BUT not able to display "Balanced"

Please help to where exactly I am getting wrong in my code

Treeid   treelatitude  treelongitude  treeCondition
  1       12.33          17.22           Balanced
  2       12.33          17.22           Healthy
  3       12.33          17.33           Dieseased
  4       13.44          17.55           Healthy
  5       11.32          17.66           Imbalanced
  6       12.33          18.33           Healthy
  7       14.44          18.44           Healthy
  8       11.22          17.22           Balanced

<?php

define('__ROOT__', dirname(dirname(__FILE__))); 
require_once(__ROOT__.'/public_html/Config.php');


 // Connecting to mysql database
    $mysqli = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_DATABASE);

 // json response array
 $response = array();


  $sq = "SELECT treeid, treelatitude, treelongitude FROM tree WHERE   (treecondition = 'Healthy','Balanced')";

 if (isset($_POST['treecondition'])) {

// receiving the post params

$treecondition = $_POST['treecondition'];

// get the tree details for google map marker
if($stmt = $mysqli->query($sq)){

   if ($stmt->num_rows) {


       while($tree = $stmt->fetch_assoc()) {

        $treeItem = array();

        $treeItem["treeid"] = $tree['treeid'];
        $treeItem["treelatitude"] = $tree['treelatitude'];
        $treeItem["treelongitude"] = $tree['treelongitude'];
        $response[] = $treeItem;

  }  echo json_encode($response);

  }

  }else {
     // user is not found with the credentials
    $response["error"] = TRUE;
   $response["error_msg"] = "Tree list view credentials are wrong. Please    try again!";
echo json_encode($response);
 }
 $mysqli->close();
 }

Answer 1

It should be WHERE treecondition IN ('Healthy','Balanced')";

EDIT

Try this:

$treecondition = $mysqli->real_escape_string($_POST['treecondition']);

"SELECT treeid, treelatitude, treelongitude FROM tree WHERE treeCondition='{$treecondition}'"

We have used real_escape_string to escape the characters posted directly. This would prevent SQL injection.

Also MySQL is case sensitive please use treeCondition instead of treecondition