在 HTML 下拉菜单中显示 PHP 变量答案

【问题标题】：Displaying PHP variable in HTML dropdown menu在 HTML 下拉菜单中显示 PHP 变量
【发布时间】：2023-04-03 14:19:01
【问题描述】：

我有一个装有安全问题和答案的 SQL 数据库。我需要在 html 下拉列表中显示它们。我以为我可以用一个while循环填充下拉字段，但我无法让它工作。

知道这有什么问题吗？

<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
</head>

<body>

<?php

$conn = mysql_connect("localhost","root","")or die (mysql_error());

mysql_select_db("assignment_3", $conn);

$data = "select securityQuestion, securityAnswer from securityquestiontable";

$result = mysql_query($data, $conn) or die (mysql_error());

while ($row = mysql_fetch_assoc($result)) {
print "<select>";
print "<option value=""";
print "disabled="disabled" selected="selected">";
print "Please Select your Undergraduate School";
print "</option>";
print "<option value="1">";
print $row["securityQuestion"];
print "</option>";
print "<option value="2">";
print $row["securityAnswer"];
print "</option>";
print "</select>";
}

?>

</body>
</html>

【问题讨论】：

您是否从 phpmyadmin 中的该查询中获得任何行？

标签： javascript php html sql

【解决方案1】：

您需要在循环之外打印语句。每次循环进行时，您都在创建一个新菜单。试试这个：

print "<select>";
while ($row = mysql_fetch_assoc($result)) {
print "<option value=""";
print "disabled="disabled" selected="selected">";
print "Please Select your Undergraduate School";
print "</option>";
print "<option value="1">";
print $row["securityQuestion"];
print "</option>";
print "<option value="2">";
print $row["securityAnswer"];
print "</option>";
}
print "</select>";

【讨论】：

【解决方案2】：

试试这样的：

echo '<select>';
while ($row = mysql_fetch_assoc($result)) {
    echo '<option value="" disabled="disabled" selected="selected">Please Select your Undergraduate School</option>';
    echo '<option value="1">'.$row["securityQuestion"].'</option>';
    echo '<option value="2">'.$row["securityAnswer"].'</option>';
}
echo '</select>';

或

print "<select>";
while ($row = mysql_fetch_assoc($result)) {
    print "<option value=""";
    print "disabled="disabled" selected="selected">";
    print "Please Select your Undergraduate School";
    print "</option>";
    print "<option value="1">";
    print $row["securityQuestion"];
    print "</option>";
    print "<option value="2">";
    print $row["securityAnswer"];
    print "</option>";
}
print "</select>";

我将选择标签放在循环之外。

【讨论】：

【解决方案3】：

你正在关闭你的打印，在任何地方都使用 "，试试这个

print "<option value='1'>".$row['securityQuestion']."</option>";

而不是

print "<option value="1">$row["securityQuestion"]</option>";

同样的，试试这个

print "<option value='' disabled='disabled' selected='selected'> Please Select your Undergraduate School </option>";

而不是

print "<option value="" disabled="disabled" selected="selected"> Please Select your Undergraduate School </option>";

并将您的选择放在循环之外。

【讨论】：

【解决方案4】：

你犯了一些小错误

请试试这个代码..

<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Untitled Document</title>
</head>

<body>

<?php

$conn = mysql_connect("localhost","root","")or die (mysql_error());

mysql_select_db("assignment_3", $conn);

$data = "select securityQuestion, securityAnswer from securityquestiontable";

$result = mysql_query($data, $conn) or die (mysql_error());
print "<select>";
while ($row = mysql_fetch_assoc($result)) {

print "<option value=""";
print "disabled="disabled" selected="selected">";
print "Please Select your Undergraduate School";
print "</option>";
print "<option value="1">";
print $row["securityQuestion"];
print "</option>";
print "<option value="2">";
print $row["securityAnswer"];
print "</option>";

}
print "</select>";
?>

</body>
</html>

【讨论】：