從數據庫只有一個數據塊如何顯示所有數據特定的一個ID

Question

只有一個數據行從數據庫中獲取。如何顯示從SELECT返回的所有數據。該SELECT應，通過調用之前抓取返回多行，查詢中的ID是外鍵的可以返回一個或多個結果

<?php 
    session_start(); 
    $r = $_SESSION["v_id"]; 
    $p = implode($r); 

    $servername = "localhost"; 
    $username = "root"; 
    $password = ""; 
    $dbname = "student"; 

    $db = new mysqli($servername, $username, $password, $dbname); 

    $query = "SELECT s_name, code FROM sponser where v_id = '$p'"; 
    $result = mysqli_query($db,$query); 
    $row = mysqli_fetch_array($result); 

    $count = mysqli_num_rows($result); 

    // fetch the all data to the database where id is v_id 
    if($count > 0) { 
     while ($row) { 
      printf ("s_name: %s code: %s", $row["s_name"], $row["code"]); 
      break; 
     } 
    } 
    ?> 

Answer 1

$query = "SELECT * FROM sponser where v_id = '$p'"; 
$result = mysqli_query($db,$query); 
while($row = mysqli_fetch_assoc($result)) { 
     //do code 
} 

Answer 2

你被扔掉你的結果集的第一行進入while循環。

行從結果集中一次一個地取出，像這樣在一個while循環中。

<?php 
session_start(); 
$r = $_SESSION["v_id"]; 
$p = implode($r); 

$servername = "localhost"; 
$username = "root"; 
$password = ""; 
$dbname = "student"; 

$db = new mysqli($servername, $username, $password, $dbname); 

$query = "SELECT s_name, code FROM sponser where v_id = '$p'"; 
$result = mysqli_query($db,$query); 

// this call was getting the first row of the result set 
// and then ignoring it as yo do nothing with it 
//$row = mysqli_fetch_array($result); 

// now you fetch one row at a time 
// each time round the while loop 
while ($row = $result->fetch_assoc()) { 
    printf ("s_name: %s code: %s", $row["s_name"], $row["code"]); 
    // the break is not required, and in fact stops 
    // the while loop after its has fetched only the first row 
    //break; 
} 

由於編寫查詢是在SQL Injection Attack 風險有一個看看發生了什麼事Little Bobby Tables即使 if you are escaping inputs, its not safe! 使用prepared parameterized statements

<?php 
session_start(); 
$r = $_SESSION["v_id"]; 
$p = implode($r); 

$servername = "localhost"; 
$username = "root"; 
$password = ""; 
$dbname = "student"; 

$db = new mysqli($servername, $username, $password, $dbname); 

$query = "SELECT s_name, code FROM sponser where v_id = ?"; 
$stmt = $db->prepare($query); 
if (!$stmt) { 
    echo $db->error(); 
    exit; 
} 
$stmt->bind_param('i', $_SESSION["v_id"]); 

$result = $db->execute(); 
if (!$stmt) { 
    echo $db->error(); 
    exit; 
} 

// Gets a result set from a prepared statement 
$result = $stmt->get_result(); 

// this fetches the result 
while ($row = $result->fetch_assoc()) { 
    printf ("s_name: %s code: %s", $row["s_name"], $row["code"]); 
} 
?>