從鏈接中的ID獲取數據庫數據時出現問題

Question

我目前有此代碼，我的目標是從數據庫中與URL中的id（blabla.php？id = 8）對應的數組（oppdrID）中獲取數據。 現在，當我運行此代碼時，什么都沒有顯示，沒有錯誤，沒有任何錯誤。

//更新的代碼。 錯誤消息“（！）致命錯誤：無法在第39行的C：\\ wamp64 \\ www \\ prosjekt \\ endre.php中將mysqli_result類型的對象用作數組”

<!doctype html>
<html>
<head>
<link rel='stylesheet' type='text/css' href='stilsett.php' />
<meta charset="utf-8">
<title>Endre oppdrag // Prosjekt - PHP</title>
</head>
<body>

<?php
include "funksjoner.inc.php";
echo "<div id='header'>";

echo navigasjon();

echo "</div>";

echo "<div id='innhold'>";

$db = kobleTil();

if (isset($_GET['oppdrID']) && is_numeric($_GET['oppdrID']) &&       $_GET['oppdrID'] > 0) {

$id = $_GET['oppdrID'];
$sql = "SELECT * FROM oppdrag WHERE oppdrID = ?";

$stmt = $db->stmt_init();
if ($stmt->prepare($sql))
$stmt->bind_param("i", $id);
$stmt->execute();

while ($nesteRad = $stmt->get_result()){

echo "<hr />";

    echo "<table id='resultat'>";
    echo "<tr><th>Navn</th><th>Type</th><th>Startdato</th><th>Sluttdato</th>  <th>Antall timer</th><th>Aktiv</th></tr>";
    echo "<tr>";
    echo "<td>" . $nesteRad['navn'] . "</td>"; 
    echo "<td>" . $nesteRad['type'] . "</td>";
    echo "<td>" . $nesteRad['startDato'] . "</td>";
    echo "<td>" . $nesteRad['sluttDato'] . "</td>";
    echo "<td>" . $nesteRad['timer'] . "</td>";
    echo "<td>" . $nesteRad['aktiv'] . "</td>";
    echo "</tr></table>";

    $stmt->close();


    echo "<hr />";

    }
}

echo    "<form action='kjoer2.php' method='post'>";
echo    "<table id='leggInn'><tr><td>
    <label for='startTid'>Starttid</label></td><td><input type='datetime-  local' name='startTid' id='skjemaLeggInn'></td></tr>
    <tr><td>
    <label for='slutTid'>Sluttid</label></td><td><input type='datetime-  local' name='slutTid' id='skjemaLeggInn'></td></tr>
    <tr><td>
    <label for='merknad'>Merknad</label></td><td><textarea name='merknad'   rows='10' cols='30'  id='skjemaLeggInn'></textarea></td></tr>
    <tr><td>
    <label for='antTimer'>Antall Timer</label></td><td><input type='text'    name='antTimer' id='skjemaLeggInn'>
    </td></tr>
    <tr><td><input type='submit' value='Legg inn'>
    </td></tr>
    </form>";

echo "</div>";
?>


</body>
</html>

Answer 1

您的網址應類似於blabla.php?oppdrID=8

然后修復代碼以獲得正確的參數，並將其添加到SQL中：

$id = $_GET['oppdrID'];
$sql = "SELECT * FROM oppdrag WHERE oppdrID = ?";

if ($statement = $db->prepare($sql)) {
    $statement->bind_param("i", $id);
    $statement->execute();
...

從鏈接中的ID獲取數據庫數據時出現問題

問題描述

1 個解決方案

解決方案1
0 2016-11-24 18:19:58

從鏈接中的ID獲取數據庫數據時出現問題

問題描述

1 個解決方案

解決方案1 0 2016-11-24 18:19:58

解決方案1
0 2016-11-24 18:19:58