[英]Displaying a php mysql result with two values with a foreach loop in PHP
I have result that I want to display as a drop down menu. 我有想要显示为下拉菜单的结果。 The query selects id and name from a table. 该查询从表中选择ID和名称。
$usersQuery = "SELECT id, name
FROM users";
$usersResult = mysqli_query ($dbc, $usersQuery);
I want to use this result as a list in a drop down menu. 我想将此结果用作下拉菜单中的列表。 This is what i have so far. 这是我到目前为止所拥有的。
<select id="dropdown" name="dropdown">
<option value="select" selected="selected">Select</option>
<?php
while ($usersRow = mysqli_fetch_array($usersResult, MYSQLI_ASSOC)){
foreach ($usersRow as $value){
echo "<option value=\"$value\"";
echo ">$value</option>\n";
}
}
?>
</select>
this would work fine if I just wanted to display name as both the value and the display to the user. 如果我只想显示name作为值和显示给用户,这将很好用。 But what I want to do is use the selected id as "value" for the select option and I want to show the name selected to the user. 但是我要做的是将所选ID用作选择选项的“值”,并且我想向用户显示所选名称。 I have tried this but it does not work. 我已经尝试过了,但是没有用。
<select id="dropdown" name="dropdown">
<option value="select" selected="selected">Select</option>
<?php
while ($usersRow = mysqli_fetch_array($usersResult, MYSQLI_ASSOC)){
foreach ($usersRow as $id=>$name){
echo "<option value=\"$id\"";
echo ">$name</option>\n";
}
}
?>
</select>
Any help would be great. 任何帮助都会很棒。
Thanks in advance. 提前致谢。
mysqli_fetch_array()
is a function which converts your query results into an array. mysqli_fetch_array()
是将查询结果转换为数组的函数。 which means you can display your values like you would with a normal array value. 这意味着您可以像使用普通数组值一样显示值。
<select id="dropdown" name="dropdown">
<option value="select" selected="selected">Select</option>
<?php
while ($usersRow = mysqli_fetch_array($usersResult, MYSQLI_ASSOC)){
echo "<option value=\"".$usersRow['id']."\"";
echo ">".$usersRow['name']."</option>\n";
}
?>
</select>
No need for the foreach
iteration; 无需foreach
迭代; mysqli_fetch_array()
already provides an associative array. mysqli_fetch_array()
已经提供了一个关联数组。 After each fetch do 每次抓取后
// Assuming "id" is a numeric value
printf('<option value="%d">%s</option>', $usersRow['id'], $usersRow['name']);
The foreach
is unnecessary - using a while loop on the mysqli_fetch_array
command will return all the results with each row in an array - you can use it like so: foreach
是不必要的-在mysqli_fetch_array
命令上使用while循环将返回数组每一行的所有结果-您可以像这样使用它:
while ($usersRow = mysqli_fetch_array($usersResult, MYSQLI_ASSOC)){
echo "<option value=\"".$usersRow['id']."\"";
echo ">".$usersRow['name']."</option>\n";
}
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