PHP - 致命錯誤：調用一個成員函數bind_param（）一個非對象 -

Question

此腳本用於與Android應用

錯誤： 2致命錯誤：調用一個成員函數bind_param（）在非對象中**在線36

在我的腳本中有三個sql查詢，第一個是正確運行，但另外兩個不是。

我真的不明白爲什麼，因爲我做的都是同樣的東西...

---

PHP

---

<?php 
ini_set("display_errors", 1); 
error_reporting(E_ALL); 

$con = mysqli_connect("******","******","******","******"); 

// Check connection 
if (mysqli_connect_errno()) { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
} 
else { 
    // If all the input are ok 
    if(isset($_REQUEST['firstName']) AND isset($_REQUEST['lastName']) AND isset($_REQUEST['phone']) AND isset($_REQUEST['linkedPhone']) AND isset($_REQUEST['imei'])) { 

     $firstname = $_REQUEST['firstName']; 
     $lastName = $_REQUEST['lastName']; 
     $phone = $_REQUEST['phone']; 
     $linkedPhone = $_REQUEST['linkedPhone']; 
     $imei = $_REQUEST['imei']; 

     $req = $con->prepare('SELECT idUser FROM user WHERE phone = ?'); 
     $req->bind_param("s", $_REQUEST['linkedPhone']); 
     $req->execute(); 
     $req -> bind_result($idFollowed); 
     $result = $req->fetch(); 
     $idFollowed = "".$idFollowed; 
     echo $idFollowed; 
     // if linkedPhone doesn't exist 
     if(!$result) { 
      $return['result'] = "Linked phone doesn't exist"; 
     } 
     else { 
      $req2 = $con->prepare('INSERT INTO user (firstName, lastName, phone, idFollowed, imei) VALUES (?, ?, ?, ?, ?)'); 
      // Work using phpMyAdmin 
      // INSERT INTO user (firstName, lastName, phone, idFollowed) VALUES ('Gael', 'Fontenelle', '', '1'); 
      $req2->bind_param("sssss", $_REQUEST['firstName'], $_REQUEST['lastName'], $_REQUEST['phone'], $idFollowed, $_REQUEST['imei']); // ERROR HERE 
      $req2->execute(); 
      // Find the id of the new user 
      $req3 = $con->prepare('SELECT idUser FROM user WHERE phone = ?'); 
      $req3->bind_param("s", $_REQUEST['phone']); // ERROR HERE 
      $req3->execute(); 
      $req3 -> bind_result($idUser); 
      $result = $req->fetch(); 
      if(!$result) { 
       $return['result'] = "Error in the registration process"; 
      } 
      else { 
       $return['result'] = "".$idUser; 
      } 
     } 
    } 
    else { 
     // Display the error 
     $return['result'] = "Missing data!"; 
    } 
    echo json_encode($return); 
} 
mysqli_close($con); 
?> 

---

SQL

---

CREATE TABLE user (
    idUser int NOT NULL auto_increment, 
    firstName varchar(50) NOT NULL, 
    lastName varchar(50) NOT NULL, 
    phone varchar(50) NOT NULL, 
    idFollowed int default 0, -- 0 if it's the primary user 
    validation boolean NOT NULL, default 0, -- 1 = VALIDATION OK 
    imei varchar(50) NOT NULL, 
    PRIMARY KEY (idUser), 
    FOREIGN KEY (idFollowed) REFERENCES idUser (user) 
); 

INSERT INTO user (firstName, lastName, phone, imei) VALUES 
    ('Steve', 'Jobs', '0836656565651', '23456789765'), 
    ('Steve', 'Wozniak', '0836656565652', '23456789765'), 
    ('Bill', 'Gates', '0836656565653', '23456789765'), 
    ('Steve', 'Balmer', '0836656565654', '23456789765'), 
    ('Larry', 'Pagen', '0836656565655', '23456789765'), 
    ('Serguei', 'Brin', '0836656565656', '23456789765'); 

Answer 1

$req2->bind_param("sssis",$_REQUEST['firstName'], $_REQUEST['lastName'], $_REQUEST['phone'], $idFollowed, $_REQUEST['imei']); 

你的第四個參數INT替換成「SSSSS」到「sssis」

Answer 2

這意味着，你的SQL查詢失敗，原因是一個錯誤。在這裏，你對如何設置連接以及$ con對象有一個問題。那麼試試這個

$con = new mysqli("myhost","myusrname","mypwd","mydb"); 

/*then check connection */ 

if($con->connection_error) 
{ 

    die("error in connecting to database"); 
} 
/*Proceed with the rest here */ 

你忘了

new 

關鍵字

您還可能在現場準備SQL錯誤。您可以檢查錯誤這樣

$resq = $con->prepare("......"); 
if(!$resq) 
{ 
    die("prepare failed ".$con->error); 
}