MySQL中的查询在PHP中失败 [英] MySQL like Query fails in PHP

问题描述

  $ result = mysql_query（SELECT * FROM MasjidMaster WHERE MasjidName LIKE（'％moh％'））或者mysql_error（）; 
  

我得到的错误是

<$ p $第24行中的/home/maximtec/public_html/masjid_folder/MasjidFinderScripts/find_by_name.php中的语法错误，意外的T_STRING
解析错误：语法错误， p>

当我在MySQL中使用它时，这个查询不起作用，但是当我把它放在一个PHP脚本中时，它不会工作
请建议一个解决方案

------------编辑：从收到的答案更改查询后-------------------- -----------------

嗯，我更新了我的查询，但现在我得到空结果

  {masjids：[{MasjidName：null，Address：null，Latitude：null，Longitude：null}] ，success：1，masjid：[]} 
  

以下是我的完整脚本： / p>

 <？php 
 
 / * 
 *以下代码将获得单个产品详细信息
 *产品由产品标识（pid）
 * / 
 
标识//数组用于JSON响应
 $ response = array（）; 
 
 
 //包含db连接类
 // require_once __DIR__。 /db_connect.php; 
 require_once dirname（__ FILE__）。 /db_connect.php ;; 
 
 //连接到db 
 $ db = new DB_CONNECT（）; 
 
 //检查发布数据
 if（isset（$ _ GET [MasjidName]））{
 $ MasjidName = $ _GET ['MasjidName']; 
 
 
 //从产品表中获取产品
 $ result = mysql_query（SELECT * FROM`MasjidMaster` WHERE`MasjidName` LIKE'％moh％'）or die （mysql_error（））; 
 
 $ response [masjids] = array（）;如果（mysql_num_rows（$ result）> 0）{
 
 
 while（$ row = mysql_fetch_array（$ result））{
 
 $ row = mysql_fetch_array（$ result）; 
 
 $ masjid = array（）; 
 $ masjid [MasjidName] = $ row [MasjidName]; 
 $ masjid [Address] = $ row [Address]; 
 $ masjid [Latitude] = $ row [Latitude]; 
 $ masjid [Longitude] = $ row [Longitude]; 
 
 //成功
 $回应[success] = 1; 
 
 //用户节点
 $ response [masjid] = array（）; 
 array_push（$ response [masjids]，$ masjid）; 
 
 // array_push（$ response [masjid]，$ masjid）; 
} 
 //回显JSON响应
 echo json_encode（$ response）; 
} else {
 //找不到产品
 $ response [success] = 0; 
 $ response [message] =未找到产品; 
 
 // echo无用户JSON 
 echo json_encode（$ response）; 
} 
} else {
 //找不到产品
 $ response [success] = 0; 
 $ response [message] =未找到产品; 
 
 // echo无用户JSON 
 echo json_encode（$ response）; 
} 
} else {
 //必填字段丢失
 $ response [success] = 0; 
 $ response [message] =必填字段丢失; 
 
 //回显JSON响应
 echo json_encode（$ response）; 
} 
？> 
  

解决方案

  $ result = mysql_query（SELECT * FROM`MasjidMaster` WHERE`MasjidName` LIKE'％moh％'）or die（mysql_error（））; 
  

稍微调整一下，以便在``中实现好的练习包名表和表列。

你也不需要（）周围（'％moh％'）

   $result = mysql_query("SELECT * FROM MasjidMaster WHERE MasjidName LIKE ('%moh%')") or die mysql_error();

The error i get is

Parse error: syntax error, unexpected T_STRING in /home/maximtec/public_html/masjid_folder/MasjidFinderScripts/find_by_name.php on line 24

This query does work when i use it in MySQL but it doesn't when I place it in a PHP Script Please suggest a solution

------------EDIT :After changing query from the received answers-------------------------------------

Well I updated my query but now I am getting null results

{"masjids":[{"MasjidName":null,"Address":null,"Latitude":null,"Longitude":null}],"success":1,"masjid":[]}

Following is my full script :

<?php

    /*
     * Following code will get single product details
     * A product is identified by product id (pid)
     */

    // array for JSON response
    $response = array();


    // include db connect class
    //require_once __DIR__ . '/db_connect.php';
    require_once dirname(__FILE__ ). '/db_connect.php';;

    // connecting to db
    $db = new DB_CONNECT();

    // check for post data
    if (isset($_GET["MasjidName"])) {
        $MasjidName = $_GET['MasjidName'];


        // get a product from products table
        $result = mysql_query("SELECT * FROM `MasjidMaster` WHERE `MasjidName` LIKE '%moh%'") or die(mysql_error());

        $response["masjids"] = array();

        if (!empty($result)) {
            // check for empty result
            if (mysql_num_rows($result) > 0) {


                while ($row = mysql_fetch_array($result)) {

                $row = mysql_fetch_array($result);

                $masjid = array();
                $masjid["MasjidName"] = $row["MasjidName"];
                $masjid["Address"] = $row["Address"];
                $masjid["Latitude"] = $row["Latitude"];
                $masjid["Longitude"] = $row["Longitude"];

                // success
                $response["success"] = 1;

                // user node
                $response["masjid"] = array();
                array_push($response["masjids"], $masjid);

    //            array_push($response["masjid"], $masjid);
                }
                // echoing JSON response
                echo json_encode($response);
            } else {
                // no product found
                $response["success"] = 0;
                $response["message"] = "No product found";

                // echo no users JSON
                echo json_encode($response);
            }
        } else {
            // no product found
            $response["success"] = 0;
            $response["message"] = "No product found";

            // echo no users JSON
            echo json_encode($response);
        }
    } else {
        // required field is missing
        $response["success"] = 0;
        $response["message"] = "Required field(s) is missing";

        // echoing JSON response
        echo json_encode($response);
    }
    ?>

解决方案

$result = mysql_query("SELECT * FROM `MasjidMaster` WHERE `MasjidName` LIKE '%moh%'") or die(mysql_error());

A little more tweaking, for good practice wrap table names and table columns in ``.

You also shouldn't need () around ('%moh%')

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