php - 将已弃用的mysql_result更改为mysqli [duplicate]

Question

这个问题已经有了答案：
MySQLi equivalent of mysql_result()?
11个答案
我正在将我不推荐的php代码mysql改成新的mysql i，但是我对这一个有问题，问题是对于新代码，我只得到了Name的回显，即第一个回显，而不是nexts（Department，Phone），从我尝试过的所有代码中，这是最接近好结果的。有什么建议吗？
我用这里找到的代码作为指南MySQLi equivalent of mysql_result()?
旧代码.php

<?php
include 'access.php';
$UNIS=$_GET['UNIS'];
mysql_select_db("Timetable") or die(mysql_error());

$query = "SELECT * FROM `Professors` WHERE `UNIS` = '$UNIS'";
$result = mysql_query($query) or die(mysql_error());
$num=mysql_num_rows($result);
$i=0;
while($i < $num) {
$Name=mysql_result($result,$i,"Name");
$Department=mysql_result($result,$i,"Department");
$Phone=mysql_result($result,$i,"Phone");
++$i;
}
echo "Name: $Name<br>";
echo "Department: $Department<br>";
echo "Phone: $Phone";
?>

new-code.php新代码

<?php include 'access.php';
$UNIS=$_GET['UNIS'];
$result = mysqli_query($link, "SELECT * FROM `Professors` WHERE `UNIS` = '$UNIS'") or die(mysqli_error());
$num=mysqli_num_rows($result);
$i=0;
while($i < $num) {
$Name= $result->fetch_assoc()["Name"];
$Department= $result->fetch_assoc()["Department"];
$Phone= $result->fetch_assoc()["Phone"];
++$i;
}
echo "Name: $Name<br>";
echo "Department: $Department<br>";
echo "Phone: $Phone";
$link->close();
?>

Answer 1

改变这个

$result = mysqli_query($link, "SELECT * FROM `Professors` WHERE `UNIS` = '$UNIS'") or die(mysqli_error());

为了这个

$result = mysqli_query($link, "SELECT * FROM Professors WHERE UNIS = '$UNIS'") or die(mysqli_error($link));

mysqli_error()应该有连接 mysqli_error($link)
PHP mysqli_error() Function
while循环中传入了错误的参数

$i=0;
while($i < $num) {

最终格式正确的代码是

$UNIS=$_GET['UNIS'];
$result = mysqli_query($link, "SELECT * FROM Professors WHERE UNIS = '$UNIS'") or die(mysqli_error($link));
$num=mysqli_num_rows($result);
if(empty($num))
{
    echo 'Empty data';
}
else
{
    $row=mysqli_fetch_array($result,MYSQLI_ASSOC);
    echo "Name: ".$row['Name']."<br>";
    echo "Department: ".$row['Department']."<br>";
    echo "Phone: ".$row['Phone']."<br>";
}

$link->close();