MySQL 选择所有 [第一个字母以 B 开头]
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Select all where [first letter starts with B]
提问by user1333327
This is a follow up question to my previous one. I want to write a MYSQL statement that echoes every entry that starts with letter B.
这是我上一个问题的后续问题。我想写一个 MYSQL 语句来回显以字母 B 开头的每个条目。
Function.php
函数.php
function getCategory() {
$query = mysql_query("SELECT author FROM lyrics WHERE author [starts with letter B]") or die(mysql_error());
while ($row = mysql_fetch_assoc($query)) { ?>
<p><a href="##"><?= $row['author']; ?></a></p>
<?php }
Category.php?category=b
Category.php?category=b
<?php include 'includes/header.php' ?>
<?php getCategory(); ?>
<?php include 'includes/footer.php' ?>
Like that I guess. And then one for every letter of the alphabet, and one with misc (numbers etc)
我猜是这样的。然后是字母表中的每个字母一个，一个带有其他（数字等）
回答by ceejayoz
SELECT author FROM lyrics WHERE author LIKE 'B%';
Make sure you have an index on author, though!
不过，请确保您有一个索引author！
回答by Bhanu Joshi
This will work for MYSQL
这将适用于 MYSQL
SELECT Name FROM Employees WHERE Name REGEXP '^[B].*$'
In this REGEXP stands for regular expression
在这个 REGEXP 代表正则表达式
and
和
this is for T-SQL
这是用于 T-SQL
SELECT Name FROM Employees WHERE Name LIKE '[B]%'
回答by Waseem Ahmad
SQL Statement:
SQL语句：
SELECT * FROM employee WHERE employeeName LIKE 'A%';
Result:
结果：
Number of Records: 4
employeeID employeeName employeeName Address City PostalCode Country
1 Alam Wipro Delhi Delhi 11005 India
2 Aditya Wipro Delhi Delhi 11005 India
3 Alok HCL Delhi Delhi 11005 India
4 Ashok IBM Delhi Delhi 11005 India
回答by sqluck
You can use:
您可以使用：
WHERE LEFT (name_field, 1) = 'B';
回答by Taz
Following your comment posted to ceejayoz's answer, two things are messed up a litte:
在您对 ceejayoz 的回答发表评论之后，有两件事被搞砸了：
$firstis not an array, it's a string. Replace$first = $first[0] . "%"by$first .= "%". Just for simplicity. (PHP string operators)The string being compared with
LIKEoperator should be quoted. ReplaceLIKE ".$first."")byLIKE '".$first."'"). (MySQL String Comparison Functions)
$first不是数组，而是字符串。替换$first = $first[0] . "%"为$first .= "%"。只是为了简单。（PHP 字符串运算符）与
LIKE运算符比较的字符串应该被引用。替换LIKE ".$first."")为LIKE '".$first."'")。（MySQL 字符串比较函数）
