php - Mysqli准备语句图像获取错误

Question

简单地，从 mysqli 中的路径搜索和影响图像，就像这样正确工作:

<?php

$DBServer = 'localhost'; // e.g 'localhost' or '192.168.1.100'
$DBUser   = 'root';
$DBPass   = '';
$DBName   = 'water';
$conn = new mysqli($DBServer, $DBUser, $DBPass, $DBName);

// check connection
if ($conn->connect_error) {
  trigger_error('Database connection failed: '  . $conn->connect_error, E_USER_ERROR);
}

$image = $GET['image'];

$sql = " SELECT kiti FROM `database` WHERE value LIKE '%$image%' ";

if(!$result = $conn->query($sql)){
    die('There was an error running the query [' . $conn->error . ']');
}
$file_path = 'photos/';

while($row = $result->fetch_assoc()){
    $src=$file_path.'/'.$row["key"];

     echo '<img class="myimg" src="'.$src.'" alt="There ya go" width="100" height="100" />';
   }

$conn->close();

  ?>

但是由于担心sql注入(inject)，像准备好的语句一样创建了它，但它没有获取图像，请帮助:

<?php

$DBServer = 'localhost'; // e.g 'localhost' or '192.168.1.100'
$DBUser   = 'root';
$DBPass   = '';
$DBName   = 'water';
$conn = new mysqli($DBServer, $DBUser, $DBPass, $DBName);

// check connection
if ($conn->connect_error) {
  trigger_error('Database connection failed: '  . $conn->connect_error, E_USER_ERROR);
}

$image = $GET['image'];

if ($stmt = $conn->prepare("SELECT kiti FROM `database` WHERE value=?")) {

    // Bind a variable to the parameter as a string. 
    $stmt->bind_param("s", $image);

    // Execute the statement.
    $stmt->execute();
    $stmt->bind_result($key);
   // Fetch the data.
    $stmt->fetch();

   $result = $stmt->get_result();

   $file_path = 'photos/';
if ($stmt->num_rows >= "1") {
   while($row = $result->fetch_assoc()){
    $src=$file_path.'/'.$row["key"];

     echo '<img class="myimg" src="'.$src.'" alt="There ya go" width="100" height="100" />';
   }
 }else{

   echo "0 records found";
}
 // Close the prepared statement.
    $stmt->close();
    $conn->close();
}




}


 ?>

我尝试了很多，阅读了更多似乎没有错误但无法指出的帖子。

Answer 1

您应该添加更多错误处理。每个方法调用都可能失败，您的脚本应该对这种情况使用react。
例如。没有 else 分支

if ($stmt = $conn->prepare("SELECT key FROM `database` WHERE value=?")) {

所以，如果准备失败你永远不会知道。我的建议是(默认情况下)首先执行失败分支。

$stmt = $conn->prepare("SELECT key FROM `database` WHERE value=?")
if (!$stmt) {
    someErrorHandlingHere();
}
else if ( !$stmt->bind_param("s", $image) ) {
    someErrorHandlingHere();
}
else if ( !$stmt->execute() ) {
    someErrorHandlingHere();
}
else if ( !$stmt->bind_result($key) ) {
    someErrorHandlingHere();
}
else ...

我想知道你的“原始”代码如何工作，因为key是mysql中的保留字/关键字，请参阅https://dev.mysql.com/doc/refman/5.5/en/keywords.html