PHP 脚本没有返回结果

Question

我有一个脚本，我尝试从数据库中获取日期。

脚本需要显示:数据库中的日期是(日期)。点击此处继续。当我在 phpMyAdmin 中运行 SQL 查询时，SQL 查询返回日期。当我在脚本中运行它时，我没有得到任何结果。

这是我的脚本:

<?php
 $servername = "localhost";
 $username = "root";
 $password = "";
 $dbname = "db";

 // Create connection
 $conn = new mysqli($servername, $username, $password, $dbname);
 // Check connection
 if ($conn->connect_error) {
  die("Connection failed: " . $conn->connect_error);
 } 

 $sql = "set lc_time_names = 'nl_NL';";
 $sql = "SELECT date_format(date, '%e %M %Y') AS date FROM table WHERE id='1'";
 $result = $conn->multi_query($sql);

 if ($result->num_rows > 0) {
  // output data of each row
  while($row = $result->fetch_assoc()) {
   echo "The date in the database is:";
   echo " " . $row['date'] . ". ";
   echo "Click here to continue.";
   }
  } else {
   echo "0";
  }
?>

当我运行此脚本时，我得到0。当我将 echo "0"; 更改为 echo ""时。 $行['日期'] 。 ". "; 我得到一个空白页面。

我做错了什么？我该如何解决这个问题？

Answer 1

我还有一项未经批准的编辑...这有效吗？

只需将您的查询/变量分成两个单独的查询/变量 - 只需担心第一个值可能不会为您保留。

像这样:

 <?php
 $servername = "localhost";
 $username = "root";
 $password = "";
 $dbname = "db";

 // Create connection
  $conn = new mysqli($servername, $username, $password, $dbname);
 // Check connection
  if ($conn->connect_error) {
  die("Connection failed: " . $conn->connect_error);
 }  

$sql = "set lc_time_names = 'nl_NL'";
$sql2 = "SELECT date_format(date, '%e %M %Y') AS date FROM table WHERE            id='1'";
 $resultZ = $conn->query($sql);
 $result = $conn->query($sql2);
 if ($result->num_rows > 0) {
  // output data of each row
  while($row = $result->fetch_assoc()) {
   echo "The date in the database is:";
    echo " " . $row['date'] . ". ";
    echo "Click here to continue.";
   }
   } else {
    echo "0";
   }
    ?>