Question

这是对我先前的问题的后续问题。我想写一份MYSQL声明,与从B封信起的每一条目一致。

职能.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 }

类别:

<?php include  includes/header.php  ?>
<?php getCategory(); ?>
<?php include  includes/footer.php  ?>

同我一样。然后,每 letter一 letter,一 letter误(编号等)

Answer 1

SELECT author FROM lyrics WHERE author LIKE  B% ;

保证您有<条码>的<>正本/条码>索引,但!

Answer 2

这项工作将在MYSQL中进行。

SELECT Name FROM Employees WHERE Name REGEXP  ^[B].*$

在这一区域进程中,定期表达

以及

这是T-SQL

SELECT Name FROM Employees WHERE Name LIKE  [B]%

Answer 3

您可以使用:

WHERE LEFT (name_field, 1) =  B ;

Answer 4

声明:

 SELECT * FROM employee WHERE employeeName LIKE  A% ;

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

Answer 5

在你发表评论后,有两点是:

缩略语 http://www.un.org/Depts/DGACM/index_french.htm Just for simplicity. (PHP string营运人 )

插图应与LIKE的操作者相比较。 http://dev.mysql.com/doc/refman/5.0/en/string-comparison-Functions.html“rel=” 缩略语

Answer 6

你也可以尝试

SELECT author FROM lyrics WHERE author LIKE BINARY  N%  ORDER BY id ASC

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