爲什麼iam在error_log文件中得到與bind_result相關的錯誤以及如何解決？

Question

蔭收到此錯誤

[07月 - 2017年11點48分47秒UTC] PHP的警告：mysqli_stmt :: bind_result（）： 綁定變量的數量不匹配字段數準備 聲明

$stmt = $con->prepare("SELECT * FROM table where Id =?"); 
$stmt->bind_param("s", $_POST['Id']); 
$stmt->execute(); 
$result = $stmt-> bind_result($Id); 
$numRows = $result->num_rows; 
if($numRows > 0) { 
if($row = $result->fetch_assoc()) 
{ 
$Taxname=$row['TaxName']; 
$Tid=$row['Id']; 
}} 

Answer 1

您需要修改您的查詢。而不是*您需要選擇從數據庫中挑選出您實際需要的數據。
例如，如果表table有列Id,TaxName那麼你將執行就像這樣：

<?php 
$sqlData = array(); 
$stmt = $con->prepare("SELECT Id,TaxName FROM table where Id =?"); 
$stmt->bind_param("i", $_POST['Id']); //id is an integer? this should be i instead of s 
$stmt->execute(); 
$stmt->store_result(); //store the result, you missed this 
$stmt-> bind_result($Id,$TaxName); //bind_result grabs the results and stores in a variable 
$numRows = $stmt->num_rows; //see the correction I made here 
if($numRows >0){ 
     while ($stmt->fetch()) { //propper way of looping thru the result set 
      $sqlData [] = array($Id,$TaxName); 
      //assoc array would look like: 
      //$sqlData [] = array("Id" =>$Id,"TaxName" => $TaxName); 
     } 
} 
$stmt->close(); //close the connection 
?> 

然後，你將有你與mysqli的查詢結束後，您可以使用結果的數組。 希望這可以幫助你更多地理解它。

Answer 2

$stmt = $con->prepare("SELECT Id,TaxName FROM table where Id =?"); 
$stmt->bind_param("s", $_POST['Id']); 
$stmt->execute(); 
$result = $stmt-> bind_result($Id,$TaxName); 
$stmt->store_result(); 
$numRows = $stmt->num_rows; 
if($numRows > 0) { 

    while($result->fetch_assoc()) 
    { 
    $newid = $Id; 
    $newtaxname= $TaxName; 
    } 
    print_r($newid)."<br>"; 
    print_r($newtaxname); 
} 

這個代碼將會給你答案沒有任何警告。

參考：http://php.net/manual/en/mysqli-stmt.bind-result.php

mysqli_stmt :: bind_result - 綁定變量到準備好的語句 結果存儲