mysql - 在一个查询中从多个表中选择记录

我有4个彼此不相关的表。我想使用一个查询而不是4个查询从这些表中获取很少的记录。

目前，这是通过4个查询完成的。码：

<?php

$result1 = $mysqli->query("SELECT val1, val2 FROM `table1` WHERE id='1'") or die(mysqli_error());
    $row1 = mysqli_fetch_array($result1, MYSQLI_ASSOC);
    $val1 = $row1['val1'];
    $val2 = $row1['val2'];

$result2 = $mysqli->query("SELECT val3, val4 FROM `table2` WHERE id='2'") or die(mysqli_error());
    $row2 = mysqli_fetch_array($result2, MYSQLI_ASSOC);
    $val3 = $row2['val3'];
    $val4 = $row2['val4'];

$result3 = $mysqli->query("SELECT val5, val6 FROM `table3` WHERE id='3'") or die(mysqli_error());
    $row3 = mysqli_fetch_array($result3, MYSQLI_ASSOC);
    $val5 = $row3['val5'];
    $val6 = $row3['val6'];

$result4 = $mysqli->query("SELECT val7, val8 FROM `table4` WHERE id='4'") or die(mysqli_error());
    $row4 = mysqli_fetch_array($result4, MYSQLI_ASSOC);
    $val7 = $row4['val7'];
    $val8 = $row4['val8'];

?>

和我在一次查询中失败的方法。码：

<?php
$resultglobal = $mysqli->query("
SELECT
tab1.val1 AS val1,
tab1.val2 AS val2,
tab2.val3 AS Val3,
tab2.val4 AS val4,
tab3.val5 AS val5,
tab3.val6 AS val6,
tab4.val7 AS val7,
tab4.val8 AS val8

FROM

table1 as tab1,
table2 as tab2,
table3 as tab3,
table4 as tab4

WHERE

tab1.id='1',
tab2.id='2',
tab3.id='3',
tab4.id='4'
") or die(mysqli_error());

$rowg = mysqli_fetch_array($resultglobal, MYSQLI_ASSOC);
$val1 = $rowg['val1'];
$val2 = $rowg['val2'];
$val3 = $rowg['val3'];
$val4 = $rowg['val4'];
$val5 = $rowg['val5'];
$val6 = $rowg['val6'];
$val7 = $rowg['val7'];
$val8 = $rowg['val8'];
?>

怎么了？这里有什么解决方案？任何建议高度赞赏。提前非常感谢您。

最佳答案

您只需要在,子句中用AND替换WHERE。
像这样：

WHERE
tab1.id='1' AND
tab2.id='2' AND
...

只要ID是唯一的（例如主键），它就可以正常工作。这是一个隐式的CROSS JOIN。

关于mysql - 在一个查询中从多个表中选择记录，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/45538808/