php - 无法将正确的结果输出到变量

Question

这个问题在这里已经有了答案:





mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc... expects parameter 1 to be resource

(31 个回答)


7年前关闭。




现在我试图从 mysql 查询将结果附加到 php 中的变量，这是使用的代码:

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

$result = mysql_query("SELECT Snip FROM snips WHERE ID='1'");

$row = mysql_fetch_array($result);

$snip = $row['Snip'];
$snip = str_replace('<?php', '&lt;?php', $snip);
$snip = str_replace('?>', '?&gt;', $snip);
?>
    <div id="editor"><?php echo $snip; ?></div>

当我运行它返回的代码时:mysql_fetch_array() 期望参数 1 是资源，在第 170 行的 C:\xampp\htdocs\home.php 中给出的 bool 值

第 170 行是:

$row = mysql_fetch_array($result);

Answer 1

你在做mysql_query创建 mysqli 后资源

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

$result = mysqli_query($con,"SELECT Snip FROM snips WHERE ID='1'");

$row = mysqli_fetch_array($result);

$snip = $row['Snip'];
$snip = str_replace('<?php', '&lt;?php', $snip);
$snip = str_replace('?>', '?&gt;', $snip);
?>
    <div id="editor"><?php echo $snip; ?></div>