無效的參數編號：綁定變量的數量不匹配令牌的數量'

Question

我有一個查詢，填充$row["app_id"]和當我通過div id="reveal"app_id更改。我想要做的是在$ revealstmt查詢中插入app_id。我不斷收到錯誤。

Fatal error: Uncaught exception 'PDOException' with message 'SQLSTATE[HY093]: Invalid  
parameter number: number of bound variables does not match number of tokens 

我的代碼如下

<?php 
    $revealid = $row["app_id"]; 
    while (isset($revealid)){ 
      $query3 = ""; 
      $revealstmt = $conn->prepare("SELECT logs.time_entry AS logs_entry, logs.description AS logs_description, substates.name AS reveal_substates_name FROM logs LEFT OUTER JOIN applications ON logs.fk_app_id = applications.pk_app_id LEFT OUTER JOIN substates ON logs.fk_substate_id = substates.pk_substate_id WHERE logs.fk_app_id = <?php echo $revealid ?> and time_entry >= curdate() - INTERVAL DAYOFWEEK(curdate()) + 14 DAY ORDER BY time_entry DESC;"); 
      $revealstmt->execute(array('query3' => $query3)); 
?> 

<div id="<?php echo $row["app_name"];?>" class='reveal-modal'> 
    <div id="reveal"> 
      <h1><?php echo $row["app_name"]; echo $revealid; ?> 
      <br /> 
      </h1> 
      <div class="accordion"> 
        <?php while ($revealrow = $revealstmt->fetch()){?> 
          <h3><a href=""><?php echo $revealrow["logs_entry"]?> -|- <? 
php echo $revealrow["reveal_substates_name"]; ?></a></h3> 
          <div><?php echo $revealrow["logs_description"]; ?></div> 
        <?php } ?> <!-- Close Tag for $revealrow --> 
    <?php } ?> <!-- Close tag for the while(isset) --> 

我一直在使用的foreach嘗試，但不能作爲$row["app_id"]不是一個數組它每次只返回1號。

希望我已經提供了所有需要的信息，如果不是，當然會添加我需要的任何信息。再次感謝社區爲您提供的所有幫助！

Answer 1

您應該在查詢中使用連接。此外，你不需要太長的查詢。以同樣的方式，你不會把所有的PHP寫在一行中，對於你的MySQL也是一樣：這是另一種語言。

$revealstmt = $conn->prepare(
    "SELECT logs.time_entry AS logs_entry, " . 
    "logs.description AS logs_description, " . 
    "substates.name AS reveal_substates_name " . 
    "FROM logs LEFT OUTER JOIN applications " . 
    "ON logs.fk_app_id = applications.pk_app_id " . 
    "LEFT OUTER JOIN substates ON logs.fk_substate_id = substates.pk_substate_id " . 
    "WHERE logs.fk_app_id = " . $revealid . " AND " . 
    "time_entry >= curdate() - INTERVAL DAYOFWEEK(curdate()) + 14 DAY " . 
    "ORDER BY time_entry DESC;"); 
$revealstmt->execute(); 

另一個問題是，你試圖將query3分配給東西。您的查詢中沒有?佔位符，因此，無需在​​中添加任何內容。但是，其他工作解決方案可能不安全，具體取決於您的$revealid來自何處。你可以把它做此保證：

$revealstmt = $conn->prepare(
    "SELECT logs.time_entry AS logs_entry, " . 
    "logs.description AS logs_description, " . 
    "substates.name AS reveal_substates_name " . 
    "FROM logs LEFT OUTER JOIN applications " . 
    "ON logs.fk_app_id = applications.pk_app_id " . 
    "LEFT OUTER JOIN substates ON logs.fk_substate_id = substates.pk_substate_id " . 
    "WHERE logs.fk_app_id = ? AND " . 
    "time_entry >= curdate() - INTERVAL DAYOFWEEK(curdate()) + 14 DAY " . 
    "ORDER BY time_entry DESC;"); 
$revealstmt->execute(array($revealid)); 

Answer 2

我相信你的查詢準備語句有錯誤的語法。試試這個：

$revealstmt = $conn->prepare("SELECT logs.time_entry AS logs_entry, logs.description AS logs_description, substates.name AS reveal_substates_name FROM logs LEFT OUTER JOIN applications ON logs.fk_app_id = applications.pk_app_id LEFT OUTER JOIN substates ON logs.fk_substate_id = substates.pk_substate_id WHERE logs.fk_app_id = ".$revealid." and time_entry >= curdate() - INTERVAL DAYOFWEEK(curdate()) + 14 DAY ORDER BY time_entry DESC;"); 

如果您迴應就應該到標準輸出，而不是字符串值/查詢您正準備。

Answer 3

$revealstmt->execute(array('query3' => $query3));這是你的問題，你的查詢有沒有約束力的值，但仍你在你的execute statment發送綁定的數組。通過將結合值，像這樣

$revealstmt = $conn->prepare("SELECT logs.time_entry AS logs_entry, logs.description AS logs_description, substates.name AS reveal_substates_name FROM logs LEFT OUTER JOIN applications ON logs.fk_app_id = applications.pk_app_id LEFT OUTER JOIN substates ON logs.fk_substate_id = substates.pk_substate_id WHERE logs.fk_app_id = :query3 and time_entry >= curdate() - INTERVAL DAYOFWEEK(curdate()) + 14 DAY ORDER BY time_entry DESC;"); 

，然後你執行YOUT查詢：

試試這個

$revealstmt->execute(array('query3' => $revealid));

OR：

$revealstmt->bindValue(':query3', $revealid);