PHP数组数据库行

Question

我在一个数据库中有 4 个表，每个表都有相同的字段名称。我想显示我认为可以通过数组完成的学生的当前年龄。我正在寻找的输出是 14，但我目前得到的输出是 11。我拥有的代码是......

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$query = "SELECT * FROM year8_records WHERE UPN = '$upn1'";
$result = mysqli_query($conn,$query) or die("Error".mysqli_error($conn));
$row = mysqli_fetch_assoc($result);

$ageyears1 = $row['age_in_years']; // value is 11 in database


$query1 = "SELECT * FROM year9_records WHERE UPN = '$upn1'";
$result1 = mysqli_query($conn,$query1) or die("Error".mysqli_error($conn));
$row1 = mysqli_fetch_assoc($result1);

$ageyears2 = $row['age_in_years']; // value is 12 in database 


$query2 = "SELECT * FROM year10_records WHERE UPN = '$upn1'"; 
$result2 = mysqli_query($conn,$query2) or die("Error".mysqli_error($conn));
$row2 = mysqli_fetch_assoc($result2);

$ageyears3 = $row['age_in_years']; // value is 13 in database


$query3 = "SELECT * FROM year11_records WHERE UPN = '$upn1'";
$result3 = mysqli_query($conn,$query3) or die("Error".mysqli_error($conn));
$row3 = mysqli_fetch_assoc($result3);

$ageyears4 = $row['age_in_years']; // value is 14 in database



$result = array($ageyears1, $ageyears2, $ageyears3, $ageyears4);

echo "<b>Current Age</b></br>" . end($result) . " Years";

当我写作时..

$result = array("11","12","13","14");

echo "<b>Current Age</b></br>" . end($result) . " Years"; 

我得到 14。 但如果可能的话，我真的需要它来自变量......

Answer 1

复制/粘贴每次都能帮到您。只是一些错别字！

$query = "SELECT * FROM year8_records WHERE UPN = '$upn1'";
$result = mysqli_query($conn,$query) or die("Error".mysqli_error($conn));
$row = mysqli_fetch_assoc($result);

$ageyears1 = $row['age_in_years']; // value is 11 in database


$query1 = "SELECT * FROM year9_records WHERE UPN = '$upn1'";
$result1 = mysqli_query($conn,$query1) or die("Error".mysqli_error($conn));
$row1 = mysqli_fetch_assoc($result1);

$ageyears2 = $row1['age_in_years']; // <-- changed


$query2 = "SELECT * FROM year10_records WHERE UPN = '$upn1'"; 
$result2 = mysqli_query($conn,$query2) or die("Error".mysqli_error($conn));
$row2 = mysqli_fetch_assoc($result2);

$ageyears3 = $row2['age_in_years']; // <-- changed

$query3 = "SELECT * FROM year11_records WHERE UPN = '$upn1'";
$result3 = mysqli_query($conn,$query3) or die("Error".mysqli_error($conn));
$row3 = mysqli_fetch_assoc($result3);

$ageyears4 = $row3['age_in_years']; // <-- changed



$result = array($ageyears1, $ageyears2, $ageyears3, $ageyears4);

echo "<b>Current Age</b></br>" . end($result) . " Years";