MySQL數據在PHP中檢索

Question

我正在嘗試在php頁面中執行一個簡單的mySQL查詢，並且不斷收到此錯誤：“警告：mysql_fetch_array（）期望參數1為資源，布爾值在...中給出”，即使查詢返回的結果是mysql工作台。 這是我的代碼：

<?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);
?>

我在這里做錯了什么？

Answer 1

您正在混合mysql_*和mysqli_*函數。

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

應該

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

有趣的是，您在該代碼正下方使用它們：

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

您可能希望將兩者結合起來以清理代碼：

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