試圖將表格放在查詢結果的周圍嗎？

Question

我無法在代碼中放一張表格，嘗試時需要一些幫助，但返回時出現“解析錯誤：語法錯誤，C：\\ xampp \\ htdocs \\ search_go.php中出現意外的<<”第27行”。 那我可以尋求有關如何插入表格的幫助嗎？

<?php

//capture search term and remove spaces at its both ends if the is any
$searchTerm = trim($_GET['keyname']);

//check whether the name parsed is empty
if($searchTerm == "")
    {
    echo "Enter name you are searching for.";
    exit();
}

//database connection info
$host = "localhost"; //server
$db = "calendar"; //database name
$user = "root"; //dabases user name
$pwd = ""; //password

//connecting to server and creating link to database
$link = mysqli_connect($host, $user, $pwd, $db);

//MYSQL search statement
$query = "SELECT * FROM caltbl WHERE evtDate LIKE '%$searchTerm%'";

$results = mysqli_query($link, $query);

<table>
/* check whether there were matching records in the table
by counting the number of results returned */
if(mysqli_num_rows($results) >= 1)
{
    $output = "";
    while($row = mysqli_fetch_array($results))
    {
        <tr>
        $output .= "date: " . $row['evtDate'] . "<br />";
        $output .= "Name: " . $row['patient'] . "<br />";
        $output .= "Course: " . $row['patientId'] . "<br />";

    }
    echo $output;
}
else
    echo "There was no matching record for the name " . $searchTerm;
?>

Answer 1

您不能只在HTML代碼中插入HTML標記：

但是，您可以使用echo直接將其發送出去：

echo "<table>";

while($row = mysqli_fetch_array($results))
{
    $output = "<tr>";
    // <tr> This is the problem line.
    $output .= "<tr>";

    $output .= "<td>date: " . $row['evtDate'] . "<br /></td>";
    $output .= "<td>Name: " . $row['patient'] . "<br /></td>";
    $output .= "<td>Course: " . $row['patientId'] . "<br /></td>";
    $output .= "</tr>";
    echo $output;
}

另外，您沒有關閉<tr> 。 我添加了一些額外的代碼片段，使每個字段都成為表中的TD，然后關閉該行。