Ajax请求后PHP中Mysqli查询无法正常运行问题求助

Hey Edoardo, let's break down the issues in your code and figure out why the query isn't working via Ajax.

1. Incorrect Check for SELECT Query Success #

Your condition if ($conn->query($sqlselect) === TRUE) is wrong. For SELECT statements, query() returns a mysqli_result object when successful, not the boolean TRUE (that's only for write operations like INSERT/UPDATE). So this condition will never be true, and all the code inside the block won't run at all.

2. SQL Injection Vulnerability #

You're directly concatenating user input variables into your SQL string, which is a huge security risk. Attackers can easily manipulate these values to execute malicious SQL commands. Always use prepared statements instead.

3. Redundant Query Execution #

You're running $conn->query($sqlselect) twice—once for the check, once to get the result. This is unnecessary and wastes database resources.

Here's how to rewrite your code to fix these issues, plus proper error handling:

// Assume $conn is your valid mysqli connection
$nome = $_POST['nome']; // Or wherever you're getting these values
$cognome = $_POST['cognome'];
$time = $_POST['time'];
$idsede = $_POST['idsede'];
$idref = $_POST['idref']; // Ensure these variables are properly initialized
$idrep = $_POST['idrep'];

// Use prepared statement to prevent SQL injection
$sqlselect = "SELECT visitatori_id FROM visitatori WHERE visitatori_nome=? AND visitatori_cognome=? AND visitatori_orastart=? AND visitatori_idsede=?";
$stmt = $conn->prepare($sqlselect);
// Bind parameters: s = string, i = integer (adjust types based on your column data types)
$stmt->bind_param("sssi", $nome, $cognome, $time, $idsede);
$stmt->execute();
$resultz = $stmt->get_result();

// Check if a matching row was found
if ($resultz->num_rows > 0) {
    $rowz = $resultz->fetch_assoc();
    $id = $rowz["visitatori_id"];
    $data = date('dmY');
    $arr = $id."/".$data.$idref.$idrep.$idsede;
    $JSON = new stdClass(); // Initialize the JSON object properly
    $JSON->value = $arr;
    // Return valid JSON response
    header('Content-Type: application/json');
    echo json_encode($JSON);
} else {
    // Handle case where no matching visitor exists
    header('Content-Type: application/json');
    echo json_encode(['error' => 'No visitor found matching the provided criteria']);
}

// Clean up resources
$stmt->close();
$conn->close();

Even with fixed PHP code, your Ajax request might fail for other reasons:

• Verify variable values: Add var_dump($nome, $cognome, $time, $idsede); exit; at the top of your PHP script to confirm you're receiving the correct values from the Ajax request.
• Enable PHP error reporting: Temporarily add error_reporting(E_ALL); ini_set('display_errors', 1); to the top of your PHP file—this will reveal any syntax or runtime errors breaking the script.
• Inspect request in browser dev tools: Go to the Network tab, find your Ajax request, and check:
  • If the correct parameters are being sent via POST/GET
  • The response status code (200 = success, 404 = wrong PHP file path, 500 = server-side error)
  • The raw response content to see if your JSON is properly formatted or if there's an error message from PHP

内容的提问来源于stack exchange，提问作者Edoardo