[英]Problems getting database data from id in link
I've currently got this code, where my goal is to fetch the data from a array in the database (oppdrID) corresponding to the id in the URL (blabla.php?id=8). 我目前有此代码,我的目标是从数据库中与URL中的id(blabla.php?id = 8)对应的数组(oppdrID)中获取数据。 Right now when I run this code nothing shows up, no errors no nothing.
现在,当我运行此代码时,什么都没有显示,没有错误,没有任何错误。
// Updated code. //更新的代码。 Error message "( ! ) Fatal error: Cannot use object of type mysqli_result as array in C:\\wamp64\\www\\prosjekt\\endre.php on line 39"
错误消息“(!)致命错误:无法在第39行的C:\\ wamp64 \\ www \\ prosjekt \\ endre.php中将mysqli_result类型的对象用作数组”
<!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>
Your URL should be something like blabla.php?oppdrID=8
您的网址应类似于
blabla.php?oppdrID=8
Then fix your code to get the right param, and add it to your SQL: 然后修复代码以获得正确的参数,并将其添加到SQL中:
$id = $_GET['oppdrID'];
$sql = "SELECT * FROM oppdrag WHERE oppdrID = ?";
if ($statement = $db->prepare($sql)) {
$statement->bind_param("i", $id);
$statement->execute();
...
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