Suppose I have this table:

假设我有这样一张表：

id | name | city
------------------
1  | n1   | c1
2  | n2   | c2
3  | n3   | c3
4  | n4   | c4

I want to check if the value c7 exists under the variable city or not.

我想检查变量City下是否存在值c7。

If it does, I will do something.

If it doesn't, I will do something else.

如果是这样的话，我会做点什么。如果没有，我会做其他的事情。

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using modern MySQLi:

使用现代MySQLi：

$mysqli = new mysqli(SERVER, DBUSER, DBPASS, DATABASE);

$city = 'c7';
$result = $mysqli->execute_query("SELECT id FROM mytable WHERE city = ? LIMIT 1", [$city]);
if($result->num_rows == 1) {
    // found
}

using legacy mysqli

使用旧版mysqli

$mysqli = new mysqli(SERVER, DBUSER, DBPASS, DATABASE);

$city = 'c7';
$stmt = $mysqli->prepare("SELECT id FROM mytable WHERE city = ? LIMIT 1");
$stmt->bind_param("s", $city);
$stmt->execute();
$result = $stmt->get_result();
if($result->num_rows == 1) {
    // found
}

using PDO:

使用PDO：

$pdo = new PDO($dsn, $user, $pass, $options);

$city = 'c7';
$stmt = $pdo->prepare("SELECT id FROM mytable WHERE city = ? LIMIT 1");
$stmt->execute([$city]);
if($stmt->rowCount() == 1) {
    // found
}

For Exact Match

"SELECT * FROM yourTable WHERE city = 'c7'"

For Pattern / Wildcard Search

"SELECT * FROM yourTable WHERE city LIKE '%c7%'"

Of course you can change '%c7%' to '%c7' or 'c7%' depending on how you want to search it. For exact match, use first query example.

当然，您可以根据您想要的搜索方式，将‘%c7%’更改为‘%c7’或‘c7%’。对于完全匹配，请使用第一个查询示例。

PHP

$result = mysql_query("SELECT * FROM yourTable WHERE city = 'c7'");
$matchFound = mysql_num_rows($result) > 0 ? 'yes' : 'no';
echo $matchFound;

You can also use if condition there.

你也可以在这里使用if条件。

This works for me :

这对我很管用：

$db = mysqli_connect('localhost', 'UserName', 'Password', 'DB_Name') or die('Not Connected');
mysqli_set_charset($db, 'utf8');

$sql = mysqli_query($db,"SELECT * FROM `mytable` WHERE city='c7'");
$sql = mysqli_fetch_assoc($sql);
$Checker = $sql['city'];

if  ($Checker != null) {
    
    echo 'Already exists';
    
} else {

    echo 'Not found';
}

SELECT
    IF city='C7'
    THEN city
    ELSE 'somethingelse'
    END as `city`
FROM `table` WHERE `city` = 'c7'

I tried to d this for a while and
$sqlcommand = 'SELECT * FROM database WHERE search="'.$searchString.'";';

$sth = $db->prepare($sqlcommand);
$sth->execute();
$record = $sth->fetch();
if ($sth->fetchColumn() > 0){}
just works if there are TWO identical entries, but,
if you replace
if ($sth->fetchColumn() > 0){}
with
if ($result){}
it works with only one matching record, hope this helps.

我尝试了一段时间，$SQLCOMMAND=‘SELECT*FROM DATABASE WHERE Search=“’.$earch String.‘”；’；$sth=$db->prepare($SQLCOMMAND)；$sth->Execute()；$Record=$sth->fetch()；if($sth->fetchColumn()>0){}只有在有两个相同的条目时才起作用，但是，如果您用if($Result){}替换if($sth->fetchColumn()>0){}，希望它只对一个匹配的记录起作用。

Assuming the connection is established and is available in global scope;

假设连接已建立并且在全局范围内可用；

//Check if a value exists in a table
function record_exists ($table, $column, $value) {
    global $connection;
    $query = "SELECT * FROM {$table} WHERE {$column} = {$value}";
    $result = mysql_query ( $query, $connection );
    if ( mysql_num_rows ( $result ) ) {
        return TRUE;
    } else {
        return FALSE;
    }
}

Usage:
Assuming that the value to be checked is stored in the variable $username;

用法：假设要检查的值存储在变量$USERNAME中；

if (record_exists ( 'employee', 'username', $username )){
    echo "Username is not available. Try something else.";
} else {
    echo "Username is available";
}

use

使用

select count(city) from mytable where city = 'c7'

This will send only a single value from the query. If it is 0 it is not present else it is present. Since you will not be use the other column values.

这将仅从查询中发送单个值。如果它是0，则它不存在，否则它存在。因为您不会使用其他列值。

For Matching the ID:

匹配ID：

Select * from table_name where 1=1

For Matching the Pattern:

若要匹配模式：

Select * from table_name column_name Like '%string%'

$result = mysqli_query($conn, "SELECT * FROM WHERE = ''";
$found = mysqli_num_rows($result) > 0 ? 'Yes' : 'no' ;
echo $found;

$RESULT=MYSQLI_QUERY($CONN，“SELECT*FROM WHERE=‘’”；$FOUND=mySQLI_NUM_ROWS($Result)>0？‘yes’：‘no’；ECHO$FOUND；

Here's is what worked with me if you're using PHP Version 8

如果您使用的是PHP版本8，下面是我使用的方法

$result = mysqli_query(database name, "SELECT id FROM mytable WHERE city = 'c7'");      
  
  if(mysqli_num_rows($result) == 0) {
       // Do Something 
  }
  else {
      // Do something else 
  }

The difference with my answer and Reindeer Wit is the 'i' in mysqli_query and mysqli_num_rows()

My Answer和Redeer Wit的不同之处在于mySQLI_QUERY和mySQLI_NUM_ROWS()中的‘I’

Also including the name of the database name inside of mysqli_query() that you are using.

还包括您正在使用的mySQLIQUERY()中的数据库名称。