PHP MySQL的:从数据库中选择并填写表格
我想从数据库中填充客户端的地址. 我使用此代码从db中进行选择:
I would like to populate addresses for client from my db. I use this code to select from db:
$stmt = $conn->prepare("SELECT * FROM peopleaddress WHERE peopleID=?");
if ( !$stmt ) {die(printf("Error: %s.\n", mysqli_stmt_error($stmt) ) );}
else if ( !$stmt->bind_param('i', $peopleID ) ) {die(printf("Error: %s.\n", mysqli_stmt_error($stmt) ) );}
else if ( !$stmt->execute() ) { die(printf("Error: %s.\n", mysqli_stmt_error($stmt) ) ); }
else {
$result = $stmt->get_result();
while($row = $result->fetch_assoc()) {
$addressID_array = array ($row['addressID']);
$addresstype_array = array ($row['addresstype']);
$addressactive_array = array ($row['active']);
$street_array = array ($row['street']);
$city_array = array ($row['city']);
$town_array = array ($row['town']);
$state_array = array ($row['state']);
$zip_array = array ($row['zip']);
$country_array = array ($row['country']);
$latitude_array = array ($row['latitude']);
$longitude_array = array ($row['longitude']);
}
} /* end else */
,此代码显示表单:
<?php
for ($i = 0; $i < count($addressID_array); $i++) {
echo '<input type="text" name="street[]" id="" placeholder="street" value="';
if (isset ($street_array[$i])){echo $street_array[$i];} echo '" />';
echo '<input type="text" name="city[]" id="city" placeholder="city" value="';
if (isset ($city_array[$i])){echo $city_array[$i]; } echo '" />';
echo '<input type="text" name="zip[]" id="zip" placeholder="postalcode" value="';
if (isset ($zip_array[$i])){echo $zip_array[$i]; } echo '" />';
echo '<input type="text" name="town[]" id="town" placeholder="town" value="';
if (isset ($town_array[$i])){echo $town_array[$i]; } echo '" />';
echo '<input type="text" name="state[]" id="state" value="';
if (isset ($state_array[$i])){echo $state_array[$i];} echo '" />';
echo '<input type="text" name="country[]" id="country" value="';
if (isset ($country_array[$i])) {echo $country_array[$i];} echo '" />';
echo '<input type="text" name="addresstype[]" id="" value="';
if (isset ($addresstype_array[$i])) {echo $addresstype_array[$i];} echo '" />';
echo '<input type="text" name="addressactive[]" id="" value="';
if (isset ($addressactive_array[$i])) {echo $addressactive_array[$i];} echo '" />'; echo '<input type="text" name="latitude[]" id="latitude" READONLY value="';
if (isset ($latitude_array[$i])) {echo $latitude_array[$i];} echo '" />';
echo '<input type="text" name="longitude[]" id="longitude" READONLY value="';
if (isset ($longitude_array[$i])) {echo $longitude_array[$i];} echo '" /> <br>';
}
?>
问题: 1)它只显示一个地址,即使在同一个客户端中有两个地址,也是如此.
Problems: 1) it only display one address, even if in db there are 2 addresses for the same client.
2)我对此很陌生.我是做对了还是有最快的(更少的代码)选项来做到这一点? 谢谢!!
2) I'm pretty new at this. Am I doing it right or there is a fastest (less code) option to do this? Thanks!!
问题在您的while循环内:
The problem is within your while loop:
$addressID_array = array ($row['addressID']);
每次while循环到变量时,这都会分配一个新数组.这些分配行都应更改为
This assigns a new array every time the while loops to the variables. These assignment lines should all be changed like
$addressID_array[] = $row['addressID'];
关于您的第二个问题:这并不是真正的答案,因为我们不知道您需要遵循哪些要求.
As for your 2nd question: it is not really answerable because we do not know the requirements you need to work against.