我想要一個 SQL Select 語句，它從數據庫中獲取值，如果有任何結果，那么在下拉菜單中顯示一個表單和結果。目前，它獲取值，但表單回顯有結果的次數。例如，如果查詢有 1 個結果，它會回顯表單一次。2 結果，回顯格式兩次，以此類推。我希望表單顯示一次，但下拉列表中包含所有結果。如果有人能指出我正確的方向，那就太好了。謝謝<?php include("conndetails.php"); $conn = new mysqli($servername, $username, $password, $dbname); $sql = "SELECT website_name FROM user_websites WHERE username='$_SESSION[user]'"; //Selects all the websites for the user that is logged in. $result = $conn->query($sql); if ($result->num_rows > 0) { while ($row = $result->fetch_assoc()) { echo '<h2>Download a website</h2> <form action="downloads.php" method="get"> <select id="website_name" name="website_name"> <option name="website_name">'. $row["website_name"]. ' </option> </select> <input type="submit" value="Download"> </form> <br> <hr> <h2>Upload to a website</h2> <form action="upload.php" method="post" enctype="multipart/form-data"> <p>Select file to upload:</p> <input type="file" name="zip_file" id="fileToUpload"> <p>Select a website to upload to:</p> <select id="website_upload_name" name="website_upload_name"> <option name="website_upload_name">'. $row["website_name"]. ' </option> </select> <br> <br> <input type="submit" value="Upload" name="submit" style="position:relative; left: -1px;"> </form> <br> <hr>'; }} ?>
2 回答
ABOUTYOU
TA貢獻1812條經驗 獲得超5個贊
創建一個變量$options并在返回數據時添加<option>到變量中。
不要將所有html代碼放在while.
if ($result->num_rows > 0) {
//Declare $options
$options = '';
while ($row = $result->fetch_assoc()) {
//Adding <option> to the var $options
$options .= '<option name="website_name">'. $row["website_name"]. '
</option>';
}
//HTML once, first part
$html = '<h2>Download a website</h2>
<form action="downloads.php" method="get">
<select id="website_name" name="website_name">';
//Adding <option> to the <select>
$html .= $options;
//HTML once, second part
$html .= '</select>
<input type="submit" value="Download">
</form>
<br>
<hr>
<h2>Upload to a website</h2>
<form action="upload.php" method="post" enctype="multipart/form-data">
<p>Select file to upload:</p>
<input type="file" name="zip_file" id="fileToUpload">
<p>Select a website to upload to:</p>
<select id="website_upload_name" name="website_upload_name">'
//Adding <option> to the second <select>
$html .= $options;
//HTML once, third part
$html .= '</select>
<br>
<br>
<input type="submit" value="Upload" name="submit" style="position:relative; left: -1px;">
</form>
<br>
<hr>';
//Printing
echo $html;
}
楊__羊羊
TA貢獻1943條經驗 獲得超7個贊
它回顯的次數與找到的結果一樣多的原因是因為您已將回顯語句放在 while 構造中。如果您希望在 if 語句的條件得到驗證時僅顯示一次回顯，則將回顯移到 while 之外，并將選項的 html 代碼放在您將在 while 內構建的變量中。在這里，我想這兩個組合都需要具有與原始代碼相同的選項：
if ($result->num_rows > 0) {
$options = '';
while ($row = $result->fetch_assoc()) {
$options .= '<option name="website_name">'. $row["website_name"]. '</option>';
}
echo '<h2>Download a website</h2>
<form action="downloads.php" method="get">
<select id="website_name" name="website_name">' . $options . '</select>
<input type="submit" value="Download">
</form>
<br>
<hr>
<h2>Upload to a website</h2>
<form action="upload.php" method="post" enctype="multipart/form-data">
<p>Select file to upload:</p>
<input type="file" name="zip_file" id="fileToUpload">
<p>Select a website to upload to:</p>
<select id="website_upload_name" name="website_upload_name">' . $options . '</select>
<br>
<br>
<input type="submit" value="Upload" name="submit" style="position:relative; left: -1px;">
</form>
<br>
<hr>';
}
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