Php MYSQL\u FETCH\u数组参数资源错误。

可能重复：

我收到以下错误消息：警告：mysql\u fetch\u array（）希望参数1是资源，布尔值在C:\xampp\htdocs中给出

它只在页面第一次加载时发生。（但如果单击按钮时表单中没有数据，通常会发生这种情况-通过javascript验证解决）

有人能告诉我我要去哪里吗？第49行出现错误，…while（$row=mysql\u fetch\u array（$result））

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您的查询在第一页加载时失败，因为您尚未提交表单。这将导致带有空

LIMIT

子句的无效语法

除非填充了

$\u POST['numberOfSequences']

，否则不要执行查询。此外，请确保已将POST字段过滤为数字，因为您的脚本容易受到当前形式的SQL注入的攻击

// Verify the field was submitted AND is a valid number
if (isset($_POST['numberofsentences']) && is_numeric($_POST['numberofsentences'])) {

  // Connect to server and select databse.
  // Note that quotes were removed from these variables. Unnecessary and bad practice 
  // to quote them unless they are part of larger interpolated strings.
  mysql_connect($host, $username, $password)or die("cannot connect");
  mysql_select_db($db_name)or die("cannot select DB");

  // Cast it to an integer
  $sn = intval($_POST['numberofsentences']);


  $query="SELECT line FROM sentence ORDER BY rand() LIMIT $sn";
  $result = mysql_query($query);

  $count = 0;

  // Verify the query succeeded...
  if ($result) {
    while ( $row = mysql_fetch_array($result) )
    { 
      // fetch result and do other stuff...
    }
 }
 // Query failed...
 else echo "An error occurred: " . mysql_error();
}

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既然不能发布空字符串，isset就足够了。empty（）是另一个有用的函数。太好了。谢谢你的帮助。现在效果很好。感谢您对语法的其他评论。

// Verify the field was submitted AND is a valid number
if (isset($_POST['numberofsentences']) && is_numeric($_POST['numberofsentences'])) {

  // Connect to server and select databse.
  // Note that quotes were removed from these variables. Unnecessary and bad practice 
  // to quote them unless they are part of larger interpolated strings.
  mysql_connect($host, $username, $password)or die("cannot connect");
  mysql_select_db($db_name)or die("cannot select DB");

  // Cast it to an integer
  $sn = intval($_POST['numberofsentences']);


  $query="SELECT line FROM sentence ORDER BY rand() LIMIT $sn";
  $result = mysql_query($query);

  $count = 0;

  // Verify the query succeeded...
  if ($result) {
    while ( $row = mysql_fetch_array($result) )
    { 
      // fetch result and do other stuff...
    }
 }
 // Query failed...
 else echo "An error occurred: " . mysql_error();
}