php

This question already has an answer here:

mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc… expects parameter 1 to be resource 31 answers

---

当你需要传递$result作为你的db结果对象时，你正在传递$query ，这是你的sql字符串 。

class DBQueries extends DBConn {
    function displayUsers(){
        $this->dbConnection();
        $query = "SELECT * FROM users";
        $result = mysql_query($query);
        while ($row = mysql_fetch_array($result)) {
            echo $row['password'];
        }
    }
}

---

你需要传递mysql_query()返回的结果句柄$result 。

---

您需要将$ result作为第一个参数传递给mysql_fetch_array函数调用，而不是$ query。

class DBQueries extends DBConn {
  function displayUsers(){
    $this->dbConnection();
    $query = "SELECT * FROM users";
    $result = mysql_query($query);
    while ($row = mysql_fetch_array($result)) {
      echo $row['password'];
    }
  }
}

链接地址: http://www.djcxy.com/p/69594.html

上一篇: 查询MySQL连接2个数据库

下一篇: PHP