Question

我需要能够检查并查看某个字符串是否在我的SQL表中的任何位置。我使用的表只有一列char。现在它说输入的所有内容都已经在表格中，即使实际上并非如此。

在SQL中，我得到的单词使用了这个：

SELECT * FROM ADDRESSES WHERE STREET LIKE '%streeetName%';

然而，在PHP中，用户输入的是单词，然后我将其存储为变量，然后试图找出一种方法来查看该变量是否在表中的某个位置。

 $duplicate = mysql_query("SELECT * FROM ADDRESSES WHERE STREET_NAME LIKE '%$streetName%'", $connect);
    if(!empty($duplicate))
    {
       echo "Sorry, only one of each address allowed.<br /><hr>";
    }

Answer 1

您需要做的不仅仅是构建查询，因为mysql_query只返回资源，而不会提供有关实际结果的任何信息。使用像mysql_num_rows这样的东西应该有用。

$duplicate = mysql_query("SELECT * FROM ADDRESSES WHERE STREET_NAME LIKE '%$streetName%'", $connect);
if(mysql_num_rows($duplicate))
{
   echo "Sorry, only one comment per person.<br /><hr>";
}

注意：不推荐使用mysql_*函数，甚至在PHP 7中将其删除。您应该使用PDO。

Answer 2

在您使用的SQL中

％streeetName％

但是在下面的查询字符串中，您使用了

％$ streeetName％

更改正确的

Answer 3

 $duplicate = mysql_query("SELECT * FROM ADDRESSES WHERE STREET_NAME LIKE '%$streetName%'", $connect);
if(!empty($duplicate))
{
   echo "Sorry, only one comment per person.<br /><hr>";
}

您需要检查

if($results->num_rows)是否有查询结果。连接和查询，检查，然后打印或错误句柄的示例，代码是松散的，不检查错误。祝你好运......

//Typically your db connect will come from an includes and/or class User...
$db = new mysqli('localhost','user','pass','database');
$sql = "SELECT * FROM `addresses` WHERE `street_name` LIKE '%$streetName%'",$connect;
//test your queries in PHPMyAdmin SQL to make sure they are properly configured.
//store the results of your query in a variable 
$results = $db->query($sql);
$stmt = '';//empty variable to hold the values of the query as it runs through the while loop
###########################################################
#check to see if you received results back from your query#
###########################################################
if($results->num_rows){
        //loop through your results and echo or assign the values as needed
        while($row = $results->fetch_assoc()){
            echo "Street Name: ".$row['STREET_NAME'];
            //define more variables from your DB query using the $row[] array.
            //concatenate values to a variable for printing in your choice further down the document.
            $address .= $row['STREET_NAME'].' '.$row['CITY'].' '$row['STATE'].' '$row['ZIP'];
        }

}else{ ERROR HANDLING }

检查PHP脚本中的SQL表中是否找到值？

3 个答案: