Use array in mysql query java

Question

I have two questions for java mysql

First, I have an array containing either 1 or 0.

If i want do to an if statement how can i then check if opt[0] = 1? is it like:

if (options[0] == 1
{
sqlBuilder.append("AND a.value1 = val[1]")
}

next question. I have another array containing different values, how can i use them in a mysql query.

Right now i got

StringBuilder sqlBuilder = new StringBuilder()
.append("SELECT * ")
.append("FROM `table1` a ")
.append("CROSS JOIN `table2` b ")
.append("CROSS JOIN ` table3` c ")
.append("CROSS JOIN `table4` d")
.append("WHERE c.`id` =" . val[0] );

So i need c.id = val[0]

For the first question; yes that looks fine aside from the missing ) after the ==1. Assuming that options[] is an int[]. — Mister
– Mister, Commented Jun 27, 2014 at 10:21
@Mister: No, he is missing a closing parenthesis, lol. Same for the second example, just replace . by + and you are fine. — gexicide
– gexicide, Commented Jun 27, 2014 at 10:22
true i miss a closing. But when i wrte if (options[0] == 1) i get the error operand strings and int.. options is a string[] — Klaus Joe Christiansen
– Klaus Joe Christiansen, Commented Jun 27, 2014 at 10:24
Shouldn't "AND a.value1 = val[1]" actually be "AND a.value1 = '" + val[1] + "'"? And btw, this is quite vulnerable to SQL injection so please use PreparedStatements instead. — Thomas
– Thomas, Commented Jun 27, 2014 at 10:26
@KlausJoeChristiansen: See my answer, you cannot compare int with String; this is neither PHP nor Javascript. — gexicide
– gexicide, Commented Jun 27, 2014 at 10:26

gexicide · Accepted Answer · 2014-06-27 10:24:14Z

2

Basically, your code is correct.

The thing that is incorrect is your Java syntax. Are you coming from PHP? Strings are concatenated with + in Java, so your last line should be

.append("WHERE c.`id` =" + val[0] );

In addition, you seem to mess up types; if your option array is a String[] then you need to compare it with strings or cast the value to int, so your first if must look like this:

if("1".equals(options[0]))

or

if(Integer.parseInt(options[0]) == 1)

Note:

As Thomas correctly states in his comment, you should consider using PreparedStatments instead of your hardcoded statements to avoid SQL injection attacks.

answered Jun 27, 2014 at 10:24

gexicide

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1 Comment

Klaus Joe Christiansen Over a year ago

Yes i am coming from PHP. I will definetly look into prepared statements. Thanks

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