Question

请尝试显示数据库中的所有内容，它会不断显示最后插入的，有用的帮助。

<?php
            include 'database/condb.php';
            $query = mysql_query("SELECT * FROM posts ");
            while($row = mysql_fetch_assoc($query)){
              $id = $row["id"];
              $username = $row["username"];
              $body =     $row["body"];
              $date_added = $row ["date_added"];
              $hasttags= $row["hashtags"];


           ?>
            <?php
          }
            ?>  

          <?php
          echo $id;
          echo $body;


          ?>

Answer 1

在while循环中打印值或将它们存储在数组中以在循环外打印它们。

    while($row = mysql_fetch_assoc($query)){
    echo $row["id"];
    echo $row["body"];
    }

Answer 2

    condb.php
    <?php
    $con = mysqli_connect("localhost","username","password","databasename");

    // Check connection
    if (mysqli_connect_errno())
      {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
      }
    ?>        

    index.php     
    <?php
        include 'database/condb.php';
        $query = "SELECT * FROM posts";
        $res = mysqli_query($con, $query);
        while($row = mysqli_fetch_assoc($res)){
            $id = $row["id"];
            $username = $row["username"];
            $body =     $row["body"];
            $date_added = $row ["date_added"];
            $hasttags= $row["hashtags"];
        echo $id;
        echo $body;
        }
    ?>

Answer 3

对于存储数据，有时使用数组会有所帮助。我通常推荐它：

<?php
    include 'database/condb.php';
    $query = mysql_query("SELECT * FROM posts");
    $array = [];
    while($row = mysql_fetch_assoc($query)){
        $id = $row["id"];
        $username = $row["username"];
        $body = $row["body"];
        $date_added = $row ["date_added"];
        $hasttags = $row["hashtags"];

        $array[$id] = ['username'=>$username, 'body'=>$body, 'date_added'=>$date_added, 'hasttags'=>$hasttags];
      }
?>

<?php
    foreach($array as $id => $value){
        echo $id; //Prints id
        echo "<br>";
        echo $value['username'].", ".$value['body'].", ".$value['date_added'].", ".$value['hasttags'];
    }
?>

或者你可以这样做：

<?php
    include 'database/condb.php';
    $query = mysql_query("SELECT * FROM posts");
    $array = [];
    while($row = mysql_fetch_assoc($query)){
        $id = $row["id"];
        $username = $row["username"];
        $body = $row["body"];
        $date_added = $row ["date_added"];
        $hasttags = $row["hashtags"];
?>

<?php
        echo $id.", ".$username.", ".$body.", ".$date_added.", ".$hasttags;
    }
?>

Answer 4

         <?php
         include 'database/condb.php';
          $query = mysql_query("SELECT * FROM posts ");
          while($row = mysql_fetch_assoc($query)){
          $id = $row["id"];
          $username = $row["username"];
          $body =     $row["body"];
          $date_added = $row ["date_added"];
          $hasttags= $row["hashtags"];

            echo $id;
             echo "<br>";
            echo $body;
       ?>
        <?php
      }
        ?>

Answer 5

作为问题的答案＆＃34;请尝试显示数据库中的所有内容，它会不断显示最后插入的，善意的帮助。＆＃34;

把echo $ id，echo $ body放在循环中。

选择所有不在mysql中工作

5 个答案: