Question

您好我在code下面使用city value来显示database -

<?php
//echo "SELECT city FROM tbl_city_master WHERE id = ".$rs->city.""
$getCityQuery = mysql_query("SELECT city FROM tbl_city_master WHERE id = ".$rs->city."");
    $resultSetCityQuery = mysql_fetch_assoc($getCityQuery);
?>
<? echo '<strong>City-</strong>' ?><? echo $resultSetCityQuery['city'];?>

所有我需要在if else条件下说没有城市选择 - 我该如何实现该代码

Answer 1

您可以使用以下代码执行此操作：

$results_count = mysql_num_rows($resultSetCityQuery);
if ($results_count > 0) {
   // do something
} else {
   echo 'No city Choosen';
}

或者您可以使用mysql_fetch_assoc并检查是否为false：

$results = mysql_fetch_assoc($resultSetCityQuery);
if ($results == false) {
  echo 'No city Choosen';
} else {
  // do somthing
}

Answer 2

if (count ($resultSetCityQuery) === 0) echo "No city choosen"
      else echo $resultSetCityQuery['city'];

或

if (empty ($resultSetCityQuery)) echo "No city choosen"
      else echo $resultSetCityQuery['city'];

Answer 3

强烈建议您停止使用mysql_ *函数，因为它们将在即将推出的php版本中弃用

您可以检查是否已设置mysql结果

if (count($resultSetCityQuery)>0) {
    echo $resultSetCityQuery['city'];
} else { 
    echo 'No city chosen';
}

Answer 4

试试这个：

foreach($resultSetCityQuery as $city){
    if(isset($city['city']) && !empty($city['city'])){
           echo "<strong>City-</strong> {$city['city']} ";
    }   
 }

Answer 5

此代码将显示可用城市

<?php
$getCityQuery = mysql_query("SELECT city FROM tbl_city_master WHERE id = ".$rs->city."");
if ($getCityQuery)
    $resultSetCityQuery = mysql_fetch_assoc($getCityQuery);

?>
<?php
if (resultSetCityQuery != null)
echo "<strong>City-</strong>$resultSetCityQuery['city']";
?>

Answer 6

尝试以下代码

    if(!!$resultSetCityQuery) {
        echo $resultSetCityQuery['city'];
    } else {
        echo "no data found";
    or 
        echo mysql_error($getCityQuery);//if there is an error ragarding to your sql statement
    }

如果在PHP中还有条件

6 个答案: