Question

我正在尝试从PHP进行查询调用。但是，我不确定为什么它不能正常工作。看起来我试图用查询做错了。在添加where子句和bindParam之前，一切正常。代码正在正确执行，然后在我进行查询和绑定后停止。有人能看出我是否正确地做到了吗？

可能与性别发布后调用有关。我无法回应$性别。

感谢您的任何见解！

 <!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <title>Practice Work 5</title>
</head>
<body>

<form action="babynames.php" method = "post">
    Year:<br>
    <input type="text" name="year">
    <input type="submit" value="Submit">
</form>

<select name = "gender">
    <option value="male">Male</option>
    <option value="female">Female</option>
</select>


</body>
</html>

<?php>
    $servername = "localhost";
$username = "root";
$password = "root";
$dbname = "baby";

$year = $_POST['year'];
$gender = $_POST['gender'];

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$sql = "SELECT Year, Name, Ranking, Gender FROM BabyNames where Year = ? and Gender = ?";
$sql -> bindParam (1, $year, PDO::PARAM_INT);
$sql -> bindParam (2, $gender, PDO::PARAM_STR);

$result = $conn->query($sql);

if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
        echo "<br> Year: ". $row["Year"]. " ; Name: ". $row["Name"]. " ; Ranking: " . $row["Ranking"] . " ; Gender: " . $row["Gender"]. " ". "<br>";
    }
} else {
    echo "0 results";
}

$conn->close();
?>

Answer 1

您的变量$ sql是字符串，而不是对象。根据{{3}} 你必须在bind params之前准备语句，如：

if ($stmt = $conn->prepare($sql)) {

    $stmt->bind_param("is", $year, $gender);
    $stmt->execute();

    $result = $stmt->get_result();
    $processedRows = 0;
    while ($row = $result->fetch_assoc()) {
        $processedRows++;
        echo "<br> Year: ". $row["Year"]. " ; Name: ". $row["Name"]. " ; Ranking: " . $row["Ranking"] . " ; Gender: " . $row["Gender"]. " ". "<br>";
    }
    if (empty($processedRows)) {  echo "0 results";  }

}
$conn->close();

Answer 2

更改此

$sql = "SELECT Year, Name, Ranking, Gender FROM BabyNames where Year == ? and Gender == ?";

到

$sql = "SELECT Year, Name, Ranking, Gender FROM BabyNames WHERE Year = ? AND Gender = ?";

PHP mySql查询调用问题

2 个答案: