Question

我有4张彼此无关的表。我想从这些表中使用一个查询而不是其中4个来获取少量记录。

目前正在进行4次查询。代码：

<?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'];

?>

和我在一个查询中创建它的失败方法。代码：

<?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'];   
?>

出了什么问题？这里有什么解决方案？任何建议高度赞赏。非常感谢你提前。

Answer 1

您只需在,子句中替换AND WITH WHERE即可。像这样：

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

只要ID是唯一的（例如主键），它就能正常工作。这是一个隐含的CROSS JOIN。

Answer 2

您可以通过这种方式进行多次查询：

$statement = "SELECT val1, val2 FROM `table1` WHERE id='1';";
$statement .= "SELECT val3, val4 FROM `table2` WHERE id='2';";
$statement .= "SELECT val5, val6 FROM `table3` WHERE id='3';";
$statement .= "SELECT val7, val8 FROM `table4` WHERE id='4'";

$values = $mysqli -> multi_query($statement) or die ('Error: '.$mysqli -> error);

while ($row = $values -> fetch_assoc()) {

    $val1 = $row['val1'];
    $val2 = $row['val2'];
    $val3 = $row['val3'];
    $val4 = $row['val4'];
    $val5 = $row['val5'];
    $val6 = $row['val6'];
    $val7 = $row['val7'];
    $val8 = $row['val8'];

}

在一个查询中从多个表中选择记录

2 个答案: