Question

如何在PHP中将爆炸词与mysql varchar进行比较？这段代码产生了我想要的东西，但它也给出了这个错误

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致命错误：在非对象上调用成员函数fetch_assoc（）......

$word = "Also, to be safe, change your password regularly... you don't have to be obsessive about it: every three hours or so should be enough. And because erring on the side of caution is always a good idea, fake your own suicide and change your identity at least once a year.";

$pieces = explode(" ", $word);

$x = 0;
while($x < word_count($word)) {  // word count function returns int (51)
  $aPiece = $pieces[$x]; // change $pieces[$x] to 'you' and then it works
  $result = $conn->query("SELECT * FROM dict WHERE english='$aPiece'");

  $z = 0;
  while($z < $num_result) // $num_result returns amount of rows in database
  {
    $row = $result->fetch_assoc(); //error line is here
    echo stripslashes($row['dutch']);
    $z++;
  }

  $x++;
}

Answer 1

我猜这个问题来自你的测试句中的don't：你忘了使用像mysql_real_escape_string这样的函数来转义引号。

例如：

$aPiece = mysql_real_escape_string($pieces[$x]);
$result = $conn->query("SELECT * FROM dict WHERE english='$aPiece'");

Answer 2

我认为

$row = $result2->fetch_assoc();

应该是

$row = $result->fetch_assoc();

因为你似乎没有任何$ result结果。

Answer 3

$ result2中的$ result2-＆gt; fetch_assoc（）;不是正确的变量，它应该是

$result->fetch_assoc();

顺便说一下，为了找到数据库中的单词，我必须删除标点符号（我认为）你的第一行应该（在爆炸之前）与此类似：

$words = strtr($words, ',.:!','');

PHP - 在PHP中将爆炸词与mysql varchar进行比较

3 个答案: