用bind_result返回超過1個mysql條目

Question

我爲我的商店創建了一個頁面，但是我使用了get_result，並且在完成了所有操作後，我意識到HostGator不允許我運行mysqlnd。所以我不得不將所有東西都轉換爲bind_result，不過我現在處於死衚衕，可能是因爲我不得不改變一切而感到沮喪......

無論如何，我的問題是我調用一些查詢來讀取表，但整個代碼在返回1個結果後停止。例如，在這段代碼中，我要求一個約會列表，儘管我在數據庫中設置了5個約會，但它只返回第一個約會。它的工作原理是從多個表中讀取，但是在1個循環處停止。如果我沒有執行$ listappointments_stmt-> free_result（），它不會讓我繼續閱讀第2個表。 首先在下一個查詢之前，所以我認爲這是我的問題，但我不知道如何工作，因爲它給了我一個布爾錯誤，如果我沒有它。任何想法歡迎！

謝謝你在前進， 的Xenos K.

<?php 
       $query="SELECT * FROM appointments"; 
       $listappointments_stmt=$mysqli->prepare($query); 
       $listappointments_stmt->execute(); 
       $listappointments_stmt->bind_result($AppId, $AppDate, $AppTime, $AppType, $AppArtist, $AppTitle, $AppClient, $AppPrice, $AppNotes); 
       while ($listappointments_stmt->fetch()) 
       { 
       $theDate=date("d-M-Y", strtotime($AppDate)); 
       echo "<tr><td>".$theDate."</td>"; 
       echo "<td>".$AppTime."</td>"; 
       $listappointments_stmt->free_result(); 
       $tempappointmentType=$AppType; 
       $query2="SELECT * FROM appointmenttypes WHERE ID=?"; 
       $listappointmentsTypes_stmt=$mysqli->prepare($query2); 
       $listappointmentsTypes_stmt->bind_param("s", $tempappointmentType); 
       $listappointmentsTypes_stmt->execute(); 
       $listappointmentsTypes_stmt->bind_result($AppTypeId, $AppTypeName, $AppTypeColor); 
       while ($listappointmentsTypes_stmt->fetch()) 
       { 
       echo "<td><span class=\"label\" style=\"background-color:".$AppTypeColor."\">".$AppTypeName."</span></td>"; 
       } 
       $listappointmentsTypes_stmt->free_result(); 
       $listappointmentsTypes_stmt->close(); 
       $tempappointmentArtist=$AppArtist; 
       $query3="SELECT * FROM staff WHERE ID=?"; 
       $listappointmentsArtist_stmt=$mysqli->prepare($query3); 
       $listappointmentsArtist_stmt->bind_param("s", $tempappointmentArtist); 
       $listappointmentsArtist_stmt->execute(); 
       $listappointmentsArtist_stmt->bind_result($ArtId, $ArtName, $ArtNickName, $ArtSurname, $ArtPhone, $ArtBirthDate, $ArtIdentificationNumber, $ArtStreetName, $ArtStreetNumber, $ArtPostalCode, $ArtCity, $ArtCountry, $ArtPosition, $ArtEmail, $ArtFacebook, $ArtInstagram); 
       while ($listappointmentsArtist_stmt->fetch()) 
       { 
       echo "<td>".$ArtName." ".$ArtNickName." ".$ArtSurname."</td>"; 
       } 
       $listappointmentsArtist_stmt->free_result(); 
       $listappointmentsArtist_stmt->close(); 
       echo "<td>".$AppTitle."</td>"; 
       $tempappointmentClient=$AppClient; 
       $query4="SELECT * FROM clients WHERE ID=?"; 
       $listappointmentsClient_stmt=$mysqli->prepare($query4); 
       $listappointmentsClient_stmt->bind_param("s", $tempappointmentClient); 
       $listappointmentsClient_stmt->execute(); 
       $listappointmentsClient_stmt->bind_result($CliId, $CliName, $CliSurname, $CliPhone, $CliBirthDate, $CliIdentificationNumber, $CliStreetName, $CliStreetNumber, $CliPostalCode, $CliCity, $CliCountry, $CliFathersFullName, $CliMothersFullName, $CliEmail, $CliFacebook, $CliInstagram, $CliNotes); 
       while ($listappointmentsClient_stmt->fetch()) 
       { 
       echo "<td>".$CliName." ".$CliSurname."</td>"; 
       echo "<td>".$CliPhone."</td>"; 
       } 
       $listappointmentsClient_stmt->free_result(); 
       $listappointmentsClient_stmt->close(); 
       echo "<td>".$AppPrice."</td>"; 
       echo "<td><a href=\"appointmentsedit.php?appointmentStatus=view&appointmentId=".$AppId."\" title=\"".$lang['view']."\"><i class=\"text-green fa fa-eye\"></i></a></td>"; 
       echo "<td><a href=\"appointmentsedit.php?appointmentStatus=edit&appointmentId=".$AppId."\" title=\"".$lang['edit']."\"><i class=\"text-blue fa fa-edit\"></i></a></td>"; 
       echo "<td><a href=\"appointmentsedit.php?appointmentStatus=delete&appointmentId=".$AppId."\" title=\"".$lang['delete']."\"><i class=\"text-red fa fa-trash-o\"></i></a></td></tr>"; 
       } 
       $listappointments_stmt->close(); 
       ?> 

Answer 1

你會是明智的，學會使用SQL語句JOIN。這將允許您僅使用一個SQL查詢來獲取所需的所有結果。與此同時，如果您在其他人中嵌套一些SQL查詢，則外部查詢（在您的案例中爲SELECT * FROM appointments）需要與其餘查詢分開的數據庫連接（在您的案例中爲$mysqli）。從連接發出一個新的查詢會重置其餘的查詢。

專業提示：避免在生產軟件中使用SELECT *。取而代之的是列出你需要的列。如果你這樣做，你的軟件將更加強大，更易於理解。