Question

我遇到mysql_fetch_array的问题。我收到警告：

mysql_fetch_array() expects parameter 1 to be resource, boolean given in D:\wamp\www\bb.php on line 6 bool(false)

这是我的代码：

if (isset($_POST['chapter'])AND isset($_POST['verse'])){
  $db=mysql_connect("localhost","fruanthony","admin");  
  mysql_select_db("gths",$db);  
  $results=mysql_query("SELECT * FROM bible where verse=".$_POST['verse']."AND            chapter=".$_POST['chapter']);  
  while($a=mysql_fetch_array($results)){    
   echo $a['info'] ;
   echo "<br>";  
  }  
}

Answer 1

$results=mysql_query("SELECT * FROM bible where verse='".$_POST['verse']."' AND            chapter='".$_POST['chapter']."'");

您缺少查询值'".$_POST['chapter']."'"中的单引号，请注意单引号

Answer 2

替换

$results=mysql_query("SELECT * FROM bible where verse=".$_POST['verse']."AND chapter=".$_POST['chapter']);

TO

$results=mysql_query("SELECT * FROM bible where verse = '".$_POST['verse']."' AND chapter = '".$_POST['chapter']."'");

Answer 3

现在尝试：）

if (isset($_POST['chapter']) AND isset($_POST['verse'])) {

// Dont ignore SQL injection
    $safeChapter = mysql_real_escape_string($_POST['chapter']);
    $safeVerse = mysql_real_escape_string($_POST['verse']);


    $db=mysql_connect("localhost","fruanthony","admin");  
    mysql_select_db("gths",$db);  
    $results=mysql_query("SELECT * FROM bible where verse='".$safeVerse."' AND chapter='".$safeChapter."';");  
    while($a=mysql_fetch_array($results))  
    {    
        echo $a['info'] ;
        echo "<br>";  
    }  
}

mysql_fetch_array发出警告

3 个答案: