php - 在 PHP 中检索 Mysql 数据

Question

我试图在 php 页面中执行一个简单的 mySQL 查询，但我不断收到此错误:“警告:mysql_fetch_array() 期望参数 1 是资源， bool 值在……”即使查询在 mysql workbench 中返回结果。这是我的代码:

<?php
$con=mysqli_connect("localhost","root","","eshkol");
// Check connection

if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

  $sql1="SET Names utf8";

 $sql = mysql_query("SELECT * FROM student WHERE idStudent=2");
    $r = mysql_fetch_array($sql);
    echo $r["idStudent"];



if (!mysqli_query($con,$sql1))
  {
  die('Error hebrew: ' . mysqli_error($con));
  }
if (!mysqli_query($con,$sql))
  {
  die('Error: ' . mysqli_error($con));
  }
echo "success";

mysqli_close($con);
?>

我在这里做错了什么？

Answer 1

您正在混合使用 mysql_* 和 mysqli_* 函数。

$sql = mysql_query($con, "SELECT * FROM student WHERE idStudent=2");
$r = mysql_fetch_array($sql);

应该是

$sql = mysqli_query($con, "SELECT * FROM student WHERE idStudent=2");
$r = mysqli_fetch_array($sql);

有趣的是您正在该代码下方使用它们:

if (!mysqli_query($con,$sql1))
  {
  die('Error hebrew: ' . mysqli_error($con));
  }
if (!mysqli_query($con,$sql))
  {
  die('Error: ' . mysqli_error($con));
  }

您可能希望将两者结合起来以清理您的代码:

$sql = mysql_query($con, "SELECT * FROM student WHERE idStudent=2");
if (!$sql) {
    die('Error: ' . mysqli_error($con));
}
$r = mysql_fetch_array($sql);