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PHP MySQL:在数据库中搜索缺少字符串(数组)的结果

[英]PHP MySQL: Search database for string (in array) missing results

我正在创建一个功能,该功能在数据库中搜索字符串。 PHP使用PHP验证程序不会显示任何错误,并且如果搜索词不存在,它将返回正确的错误。 我的问题是,在数据库中名为“并置”(当前是数据库中唯一的条目)的列中搜索“急匆匆地放弃”时,没有返回结果。 虽然我可以使用phpMyAdmin看到该条目确实存在。

用户使用以下HTML将字符串输入到输入字段中:

<form action='http://www.murkyfiles.esy.es/search.php' method='GET'>
  <center>
    <p><label for='search'>Please enter your question as accurately as possible:</label></p>
    <p><input type='search' size='90' name='search'></p>
    <p><input type='submit'  name='submit' value='Find answer'></p>
  </center>
</form>

使用以下PHP在数据库中搜索输入的术语:

<?php
$button = $_GET [ 'submit' ];
$search = $_GET [ 'search' ];
$host = "[HOST URL]";
$username = "[USERNAME]";
$password = "[PASSWORD]";
$database = "[DATABASE]";
$searchlength = strlen($search);

if( !$button )
  echo "You didn't submit a keyword";
else {
  if( strlen( $search ) <= 1 )
    echo "Search term too short";
  else {
    echo "You searched for <b> $search </b> <hr size='1' > </ br > ";

    // Connect to database

    $con = mysqli_connect ( $host, $username, $password );

    if(!$con) {
        die('Could not connect: ' .PDO::errorInfo());
    }

    mysqli_select_db ( $con, $database );

    $search = str_split($search, $searchlength);

    $construct = " SELECT * FROM 'coll_test' WHERE collocation LIKE '%$search%' ";

    $run = mysqli_query( $con, $construct );

    //Fetch and return search results.

    if ($foundnum == 0)
        echo "Sorry, there are no matching results for <b> $search[0] </b>.
        </ br >
        </ br > 1. Try presenting your Something is wrong in a more academic manner. Guidance can be found on the majority of University websites without need for registration.
        </ br > 2. Try more common words/phrases with similar meaning. This search focuses on colloquialisms - commonly used phrases within a language.
        </ br > 3. Please check your spelling";

    else {
      echo "$foundnum results found !<p>";

      while ( $runrows = mysqli_fetch_assoc($run) ) {
          $collocation = $runrows ['collocation'];
        echo "<a href='$url'> <b> $title </b> </a> <br> $desc <br> <a href='$url'> $url </a> <p>";
      }
    }
  }
}

我看过各种类似的问题,但没有一个提供解决方案。

为了澄清,数据库表的列标题如下:

搭配 | | | 长度 google-结果 | 结结果 | 雅虎结果 | 网址链接 | 维基 | 日期

到目前为止,我的数据库中只有一个条目:

collocation = abandon hastily
left = abandon
right = NULL
length = 2
google-results = 24000000
bing-results = 386000
yahoo-results = 385000
url-link = oxforddictionary.so8848.com/search1?word=abandon
wiki = 0
date = [TIMESTAMP]

结束表格表格和列名称的引号,而不是使用反引号,并且您的代码已打开以供sql注入用户prepare和bind语句阻止

$search = $_GET ['search'];// get your value
$like = "%$search%";//
$stmt = $con->prepare("SELECT `collocation`,`left`,`right` FROM `coll_test` WHERE collocation LIKE ?");
$stmt->bind_param('s', $like);
$stmt->execute();
$stmt->bind_result($collocation,$left,$right);
$rows = $stmt->num_rows;// check your query return result of not
if ($rows > 0) {

    while ($stmt->fetch()) {// fetch data from query
       printf ("%s (%s)\n", $collocation,$left,$right);
       // fetch data form result set
    }
} else {
    echo "Sorry, there are no matching results for <b> $search </b>.";
}

相当重写,带有以下注释:

$con = mysqli_connect ( $host, $username, $password, $database );

//$search = str_split($search, $searchlength); ///??? See below
$searchSafe = preg_replace("/[^0-9a-z-_ ]/i","",$search); //example only.

$construct = " SELECT COUNT(*) AS found FROM `coll_test` 
WHERE collocation LIKE '%".$searchSafe."%' ";

$run = mysqli_query($con, $construct);
$result = mysqli_fetch_array($run);

print $result['found']." number of results found!"; //for example.

1)您可以在MySQLi连接功能中包含数据库引用。

2) str_split返回一个数组,但是您将结果用作字符串。 这令人困惑且不正确,您打算如何处理?

  • $_GET['search']始终是字符串类型,因此您无需将其用作数组或任何基于数组的混乱对象。

3)使用外部函数手动返回number_rows计数可能不准确, 而是SELECT语句中使用COUNT

4)您忘记返回实际查询的结果! 因此,在上面我插入了一个mysql_fetch_array结果以查看结果数。 您也没有为$foundnum变量定义值。

5)您正在将PDO与MySQLi混合,这两种连接方法是互斥的 他们不混合。

6)您对SQL注入和数据库妥协持开放态度,您需要使用Prepared Statements(如Saty所示 ),并使用preg_replace东西(或其他REGEX解析器)从字符串中删除无效字符,例如:

$searchSafe = preg_replace("/[^0-9a-z-%_ ]/i","",$search); //example only.

上面的意思是字符串中仅允许使用0-9或az(不区分大小写, /i )或-%_

7)表名或列名( 'coll_test' )不应用单引号引起来,相反,它们应全部用反引号引起来。 在MySQL中单引号是包含数据的字符串。

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