Question

我写的代码有效，但我仍然坚持：

注意：未定义的索引：第35行的C：\ xampp \ htdocs \ portfolio \ periode4 \ graden \ array.php中的id。

有谁知道我要采取什么措施来取消通知？

<?php
function connect()
  {
    // Make a MySQL Connection
    $link = mysqli_connect("localhost", "root", "", "school") or die(mysqli_error());
    return $link;
  }


if($_GET) {
    $id = $_GET["id"];
}

// select query 
$query = "SELECT * FROM graden" ;


$result = mysqli_query(connect(),$query);   


// tabel maken
$table = "<table border='1'>";

$table .= "<tr>
             <th> ID </th>
             <th>Graden Celcius</th> 
             <th>Aanduiding</th>
             <th>Image</th> 
        </tr>";


// keeps getting the next row until there are no more to get
    while($row = mysqli_fetch_assoc( $result )) {

    $table .="<tr ".($_GET['id'] == $row['id']?" style='background-color:yellow'":"").">";

    // Print out the contents of each row into a table
    foreach ($row as $key => $value)
        {
            if ($key == "Image")
            {
                $table .="<td><img src='$value' /></td>"; 
            }
                elseif ($key == "id")
                {
                    $table .="<td><a href='array.php?id=$value'>$value</a></td>"; 
                }
                    else
                    {
                        $table .="<td>$value</td>";         
                    }

        }
        $table .= "</tr>";
        }
        $table .="</table>";

        echo $table;
    // optie lijst

//maken select query

$query = "SELECT id, temp FROM graden ";


//uitvoeren select query
$result = mysqli_query(connect(), $query) ;


    //tonen alle gegevens
    $select = "
        <form action=\"array.php\" method=\"get\">\n
            <select name=\"id\">\n
                <option value='id'>---all---</option>\n
    ";
    while($row = mysqli_fetch_assoc( $result ))
        {
            $select .= "<option value='".$row['id']."'>".$row["temp"]."</option>\n"; 
        }
            $select .= "</select>\n";
            $select .= "<input type=\"submit\">\n";
            $select .= "</form>\n";

    echo $select;

Answer 1

如果网址上没有id参数，那么当然会失败：

 $table .="<tr ". ($_GET['id'] == $row['id']?" style='background-color:yellow'":"").">";

尝试

 $table .="<tr ". ( isset($_GET['id']) && $_GET['id'] == $row['id']?" style='background-color:yellow'":"").">";

但这可能会破坏逻辑，你需要检查..

注意：未定义索引：id in

1 个答案: