mysql查詢數組計數器

Question

抱歉，如果我有錯誤的術語。

我在PHP中有一個for循環，用於操作mysql查詢...

for ($i = 0; $i <count($user_id_pc); $i++) 
{
$query2 = " SELECT job_title, job_info FROM job_description WHERE  postcode_ss = '$user_id_pc[$i]'";

$job_data = mysqli_query($dbc, $query2);

$job_results = array();

    while ($row = mysqli_fetch_array($job_data))
        {
            array_push($job_results, $row);
        }

}

當我插入...時給出的結果...

print_r ($job_results);

在屏幕上-> Array（）

例如，如果我將查詢從$user_id_pc[$i]更改$user_id_pc[14] ， $user_id_pc[14]收到一組結果。

如果我在查詢后和for循環中包含此代碼

echo $i;
echo $user_id_pc[$i] . "<br>";

我收到計數器$i所在的數字，后跟該計數器位置的數組中的數據。

我不確定為什么數組$job_results在使用計數器$i的查詢中為空，但如果我手動輸入數字則不然？

我需要逃脫嗎？

完整代碼

        <?php
print_r ($user_id_pc);

  //Select all columns to see if user has a profile
     $query = "SELECT * FROM user_profile WHERE user_id = '" . $_SESSION['user_id'] . "'";

      //If the user has an empty profile direct them to the home page

  $data = mysqli_query($dbc, $query);

  if (mysqli_num_rows($data) == 0) 
    {
        echo '<br><div class="alert alert-warning" role="alert"><h3>Your appear not to be logged on please visit the<a href="index.php"> home page</a> to log on or register. <em>Thank you.</em></h3></div>';
    }
//Select data from user and asign them to variables
    else
    {
          $data = mysqli_query($dbc, $query);
      if (mysqli_num_rows($data) == 1) 
      {
          $row = mysqli_fetch_array($data);
          $cw_job_name = $row['job_description'];
          $cw_rate = $row['hourly_rate'];
          $job_mileage = $row['mileage'];
          $job_postcode = $row['postcode'];
          $response_id = $row['user_profile_id'];
      }
    }


    for ($i = 0; $i <count($user_id_pc); $i++) 
        {
            $query2 = " SELECT job_title, job_info FROM job_description WHERE  postcode_ss = '{$user_id_pc[$i]}'";                                        
    $job_data = mysqli_query($dbc, $query2);
    $job_results = array();
        while ($row = mysqli_fetch_array($job_data))
            {
                array_push($job_results, $row);
            }

    echo $i;
?>
<br>
<?php

}
print ($query2);
print $user_id_pc[$i];
         ?>

Answer 1

這主要是語法錯誤，正確的語法應為：

$query2 = " SELECT job_title, job_info FROM job_description WHERE  postcode_ss = '{$user_id_pc[$i]}'";

請注意，這是正確的語法，但仍然是錯誤的！ 出於兩個原因，第一個原因是執行聯接或子查詢或簡單的IN（array）類型查詢比遍歷和查詢多次幾乎總是更好（更快，更高效，占用更少的資源）。

第二個問題是，以這種方式傳遞參數使您容易受到sql注入的攻擊。 您應該使用准備好的語句。

正確的方法

if(count($user_id_pc)) {
    $stmt = mysqli_stmt_prepare(" SELECT job_title, job_info FROM job_description WHERE  postcode_ss = ?");
    mysqli_stmt_bind_param($stmt, "s", "'" . implode("','",$user_id_pc) . "'");
    mysqli_stmt_execute($stmt);
}

請注意，for循環已由簡單的if代替

Answer 2

您必須檢查查詢變量，而不是：

$query2 = " SELECT job_title, job_info FROM job_description WHERE  postcode_ss = '$user_id_pc[$i]'"

你嘗試過這個嗎？

$query2 = " SELECT job_title, job_info FROM job_description WHERE  postcode_ss = '" . $user_id_pc[$i] . "' ";

另一件事，嘗試不同的方法：

while ($row = mysqli_fetch_array($job_data))
    {
        $job_results[] = array("job_title" => $row["job_title"], "job_info" => $row["job_info");
    }

然后嘗試打印值。

Answer 3

抱歉，但是我喜歡foreach（） ，所以您的工作代碼是：

<?php

 // To store the result
 $job_results = [];

 foreach($user_id_pc as $id ){
   // selecting matching rows   
   $query2 ="SELECT job_title, job_info FROM job_description WHERE postcode_ss = '".$id."'";
   $job_data = mysqli_query($dbc, $query2);
   // checking if query fetch any result
   if(mysqli_num_rows($job_data)){
      // fetching the result
      while ($row = mysqli_fetch_array($job_data)){
        // storing resulting row
        $job_results[] = $row;
     }
   }
}
// to print the result
var_dump($job_results);