Erstellen einer SELECT-Abfrage als vorbereitete MySQL-Anweisung unter Verwendung einer dynamischen Anzahl von LIKE-Bedingungen
P粉269847997
P粉269847997 2024-03-25 23:18:00
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Ich versuche, vorbereitete Anweisungen für Benutzereingaben zu schreiben. Die Anzahl der Parameter ändert sich je nach Benutzereingabe. Oam probiert diesen Code aus

PHP-Code:

$string          = "my name";
$search_exploded = explode( " ", $string );
$num             = count( $search_exploded );
$cart            = array();
for ( $i = 1; $i <= $num; $i ++ ) {
    $cart[] = 's';
}
$str          = implode( '', $cart );
$inputArray[] = &$str;
$j            = count( $search_exploded );
for ( $i = 0; $i < $j; $i ++ ) {
    $inputArray[] = &$search_exploded[ $i ];
}
print_r( $inputArray );
foreach ( $search_exploded as $search_each ) {
    $x ++;
    if ( $x == 1 ) {
        $construct .= "name LIKE %?%";
    } else {
        $construct .= " or name LIKE %?%";
    }
}
$query = "SELECT * FROM info WHERE $construct";
$stmt  = mysqli_prepare( $conn, $query );
call_user_func_array( array( $stmt, 'bind_param' ), $inputArray );
if ( mysqli_stmt_execute( $stmt ) ) {

    $result = mysqli_stmt_get_result( $stmt );
    if ( mysqli_num_rows( $result ) > 0 ) {
        echo $foundnum = mysqli_num_rows( $result );
        while( $row = mysqli_fetch_array( $result, MYSQLI_ASSOC ) ) {

            echo $id = $row['id'];
            echo $name = $row['name'];
        }
    }
}

Wenn ich print_r($inputArray) die Ausgabe ist:

Array ( [0] => ss [1] => my [2] => name )

Im Fehlerprotokoll werden keine Fehler angezeigt.

Was ist schief gelaufen?

P粉269847997
P粉269847997

Antworte allen(2)
P粉826429907

编写一个通用查询处理程序并向其传递您的查询、参数数组和参数类型列表。返回一组结果或消息。这是我自己的 mysqli 个人版本(我主要使用 PDO,但也为此设置了类似的功能)。对插入、更新和删除执行相同的操作。然后只需维护您的一个库并将其用于您所做的所有事情:) 请注意,如果您以此开始,您可能希望更好地处理连接错误等。

connect_error) {
      return false;
    }
    return $con;
}

// generic select function.
// takes a query string, an array of parameters, and a string of
// parameter types
// returns an array - 
//   if $retVal[0] is true, query was successful and returned data
//   and $revVal[1...N] contain the results as an associative array
//   if $retVal[0] is false, then $retVal[1] either contains the 
//   message "no records returned" OR it contains a mysql error message

function selectFromDB($query,$params,$paramtypes){

    // intitial return;
    $retVal[0]=false;

    // establish connection
    $con = getDBConnection();
    if(!$con){
        die("db connection error");
        exit;
    }

    // sets up a prepared statement
    $stmnt=$con->prepare($query);
    $stmnt->bind_param($paramtypes, ...$params);
    $stmnt->execute();

    // get our results
    $result=$stmnt->get_result()->fetch_all(MYSQLI_ASSOC);
    if(!$result){
    $retVal[1]="No records returned";
    }else{
        $retVal[0]=true;
        for($i=0;$iclose();

    return $retVal;

}

$myusername=$_POST['username'];
$mypassword=$_POST['password'];

// our query, using ? as positional placeholders for our parameters
$q="SELECT useridnum,username FROM users WHERE username=? and password=?";

// our parameters as an array - 
$p=array($myusername,$mypassword);

// what data types are our params? both strings in this case
$ps="ss";

// run query and get results
$result=selectFromDB($q,$p,$ps);

// no matching record OR a query error
if(!$result[0]){
    if($result[1]=="no records returned"){
        // no records
        // do stuff
    }else{
        // query error
        die($result[1]);
        exit;
    }   
}else{  // we  have matches!
    for($i=1;$i$val){
            print("key:".$key." -> value:".$val);
        }
    }
}

?>
P粉787806024

% 环绕参数,而不是占位符。

我的代码片段将使用面向对象的 mysqli 语法,而不是您的代码演示的过程语法。

首先您需要设置必要的成分:

  1. WHERE 子句表达式 -- 用 OR 分隔
  2. 值的数据类型 - 您的值是字符串,因此使用“s”
  3. 要绑定到准备好的语句的参数

我将把 #2 和 #3 组合成一个变量,以便使用 splat 运算符(...)进行更简单的“解包”。数据类型字符串必须是第一个元素,然后一个或多个元素将代表绑定值。

作为逻辑包含,如果 WHERE 子句中没有条件,则使用准备好的语句没有任何好处;直接查询表即可。

代码:(PHPize.online 演示)

$string = "Bill N_d Dave";

$conditions = [];
$parameters = [''];
foreach (array_unique(explode(' ', $string)) as $value) {
    $conditions[] = "name LIKE ?";
    $parameters[0] .= 's';
    // $value = addcslashes($value, '%_'); // if you want to make wildcards from input string literal. https://stackoverflow.com/questions/18527659/how-can-i-with-mysqli-make-a-query-with-like-and-get-all-results#comment132930420_36593020
    $parameters[] = "%{$value}%";
}
// $parameters now holds ['sss', '%Bill%', '%N_d%', '%Dave%']

$query = "SELECT * FROM info";
if ($conditions) {
    $stmt = $mysqli->prepare($query . ' WHERE ' . implode(' OR ', $conditions));
    $stmt->bind_param(...$parameters);
    $stmt->execute();
    $result = $stmt->get_result();
} else {
    $result = $conn->query($query);
}
foreach ($result as $row) {
    echo "
{$row['name']}
\n"; }

对于任何寻找类似动态查询技术的人:

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