将下拉列表从JSON填充到PHP
下面是我的JSON结果 当我想在PHP下拉列表中填充“factoryList”时,我得到了一个奇怪的结果。下面是结果 下面是我当前的php代码:将下拉列表从JSON填充到PHP,php,json,sql-server,Php,Json,Sql Server,下面是我的JSON结果 当我想在PHP下拉列表中填充“factoryList”时,我得到了一个奇怪的结果。下面是结果 下面是我当前的php代码: <!DOCTYPE html> <html lang="en"> <?php $url = 'http://172.20.0.45/TGWebService/TGWebService.asmx/factoryList'; // path to your JSON file $data = file_get_cont
<!DOCTYPE html>
<html lang="en">
<?php
$url = 'http://172.20.0.45/TGWebService/TGWebService.asmx/factoryList'; // path to your JSON file
$data = file_get_contents($url); // put the contents of the file into a variable
$characters = json_decode($data); // decode the JSON feed
foreach ($characters->factoryList as $character) {
echo '<select class="form-control" name="Fac_ID" class="form-control">';
echo "<option value='$character->facID'>$character->facName</option>";
echo "</select>";
}
?>
</html>
您需要将选择的生成移动到循环之外:
echo '<select class="form-control" name="Fac_ID" class="form-control">';
foreach ($characters->factoryList as $character) {
echo "<option value='$character->facID'>$character->facName</option>";
}
echo "</select>";
echo';
foreach($characters->factoryList作为$character){
回显“$character->facName”;
}
回声“;
请重复选择标记而不是选择标记。@RadamelFalcao无需担心-我很高兴能提供帮助。抱歉打断你,伙计,你能解决这个问题吗?