Question

我目前正在学习MySQL和php5。我目前在使用php脚本显示数据时遇到问题。我正在学习使用数据库路径并努力提出数据。

这是我的mysql表

+----+------+------------------------------+
| id | name | path                         |
+----+------+------------------------------+
|  1 | test | test                         |
|  2 | 1    | /masonry-practice/flag/1.gif |
|  3 | 2    | /masonry-practice/flag/2.gif |
+----+------+------------------------------+

这是我的PHP代码

  <?php
  $con=mysqli_connect("localhost","root","password","db");
  // Check connection
  if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

  $result = mysqli_query($con,"SELECT * FROM img");

  while($row = mysqli_fetch_array($result)) {
echo $row['name'] . " " . $row['path'];
echo "<br>";
  }

  mysqli_close($con);
  ?>

Answer 1

您必须使用它来获取值

$row = mysqli_fetch_array($result, MYSQLI_ASSOC);

执行任务的更好方法是

<?php

  $con=new mysqli(("localhost","root","password","db");
  // Check connection
  if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

$result = $con->query("SELECT * FROM img");

while($row = $result->fetch_array())
{
$rows[] = $row;
}

foreach($rows as $row)
{
echo $row['name'] . " " . $row['path'];
echo "<br>";
}

/* free result set */
$result->close();

/* close connection */
$con->close();
?>

Answer 2

试试这个并在shell上执行脚本以获得问题

<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
$conn = New mysqli(SERVER, USER, PASS, NAME);
if (!$conn)
{
die("Database connection failed: " . mysqli_error());
}
$result = $conn->query("SELECT * FROM `img`");
while($row = mysqli_fetch_assoc($result))
{
echo $row['name'] . " " . $row['path']; 
echo "<br>";
} 
?>

PHP / MySQL检索图像路径

2 个答案: