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PHP来自多个表，但在同一个数据库中(PHP results from multiple tables but in the same database)

 我有我的PHP SELECT查询工作，但我需要它显示超过1表的数据。  
$sql = "SELECT * from paymentPersonal where `custID`='$custID'";

$result = $conn->query($sql);

if ($result->num_rows > 0) {
     // output data of each row
     while($row = $result->fetch_assoc()) {
         echo "<br> custID: ". $row["custID"]. " - FirstName: ". $row["firstname"]. " LastName" . $row["lastname"] .  " - Mobile: ". $row["mobile"]. " - homephone: ". $row["homephone"]. " - Email: ". $row["email"]."<br>";
 
 这是显示1表的数据的代码，但我需要几个，我试过将它更改为此，看看它会显示更多的表：  
$sql = "SELECT * from `paymentPersonal`, `paymentsPayment` where `custID`='$custID'";

    $result = $conn->query($sql);

    if ($result->num_rows > 0) {
         // output data of each row
         while($row = $result->fetch_assoc()) {
             echo "<br> custID: ". $row["custID"]. " - FirstName: ". $row["firstname"]. " LastName" . $row["lastname"] .  " - Mobile: ". $row["mobile"]. " - homephone: ". $row["homephone"]. " - Email: ". $row["email"]."<br>";

             echo "<br> custID: ". $row["custID"]. " - CcName: ". $row["nameoncard"]. " CcNumber" . $row["ccnumber"] .  " - ccYear: ". $row["year"]. " - ccMonth: ". $row["month"]. " - ccCode: ". $row["code"]."<br>";
 
 但没有任何运气只是得到一个错误说;  
Notice: Trying to get property of non-object in /Applications/XAMPP/xamppfiles/htdocs/horizonphotography/findingcode.php on line 19
0 results
 
 （仅供参考，所有表格中的custID ）如果您对如何在其他表格上显示数据有所了解，请提供帮助！ 谢谢！ 

ive got my php SELECT query working but i need it to display more then 1 table worth of data. 
$sql = "SELECT * from paymentPersonal where `custID`='$custID'";

$result = $conn->query($sql);

if ($result->num_rows > 0) {
     // output data of each row
     while($row = $result->fetch_assoc()) {
         echo "<br> custID: ". $row["custID"]. " - FirstName: ". $row["firstname"]. " LastName" . $row["lastname"] .  " - Mobile: ". $row["mobile"]. " - homephone: ". $row["homephone"]. " - Email: ". $row["email"]."<br>";
 
this is the code that displays the data for the 1 table but i need serveral and ive tried changing it to this to see wether it would display more tables: 
$sql = "SELECT * from `paymentPersonal`, `paymentsPayment` where `custID`='$custID'";

    $result = $conn->query($sql);

    if ($result->num_rows > 0) {
         // output data of each row
         while($row = $result->fetch_assoc()) {
             echo "<br> custID: ". $row["custID"]. " - FirstName: ". $row["firstname"]. " LastName" . $row["lastname"] .  " - Mobile: ". $row["mobile"]. " - homephone: ". $row["homephone"]. " - Email: ". $row["email"]."<br>";

             echo "<br> custID: ". $row["custID"]. " - CcName: ". $row["nameoncard"]. " CcNumber" . $row["ccnumber"] .  " - ccYear: ". $row["year"]. " - ccMonth: ". $row["month"]. " - ccCode: ". $row["code"]."<br>";
 
but without any luck just getting an error saying;  
Notice: Trying to get property of non-object in /Applications/XAMPP/xamppfiles/htdocs/horizonphotography/findingcode.php on line 19
0 results
 
(FYI the custID is in all of the tables) if you have an idea on how to display the data on this other table please help! thank you!

原文：https://stackoverflow.com/questions/37022063

更新时间：2021-09-09 19:09

最满意答案

 我不熟悉croniter，但freezegun可以帮助解决这个问题 - 它使用freezegun.api.FakeDatetime实例修补对datetime.datetime所有引用，因此在some_code() ，任何datetime.datetimes都应该使用freezegun.api.FakeDatetime实例freezegun.api.FakeDatetime 。 从经验来看，如果你使用freezegun试图模拟时间的流逝，你也可以避免许多令人头疼的间歇性测试失败。  
from datetime import datetime, timedelta
from freeze_gun import freeze_time
from other_module import some_code

fake_today = datetime.datetime(2017, 7, 11)
with_freeze_time(fake_today):
    some_code()
# simulate passage of time to tomorrow
with_freeze_time(fake_today + timedelta(days=1))
    some_code()

I'm not familiar with croniter, but freezegun may help with this— it patches all references to datetime.datetime with instances of freezegun.api.FakeDatetime, so within some_code(), any datetime.datetimes should use instances of freezegun.api.FakeDatetime. Speaking from experience, you'll also save yourself lots of headaches with intermittent test failures when trying to simulate the passage of time if you use freezegun. 
from datetime import datetime, timedelta
from freeze_gun import freeze_time
from other_module import some_code

fake_today = datetime.datetime(2017, 7, 11)
with_freeze_time(fake_today):
    some_code()
# simulate passage of time to tomorrow
with_freeze_time(fake_today + timedelta(days=1))
    some_code()

PHP来自多个表，但在同一个数据库中(PHP results from multiple tables but in the same database)

最满意答案

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