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Only when the query inputted placeholders are LIKE either column, the restaurant will return as result I have attempted to create this query using this, but the syntax must be wrong as it does not return as desired;
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I have tried to use the % many times in different ways, but i have no more ideas of how to use it, just to make it work If any one could tell me what is wrong i would appreciate that
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Thanks a lot, this did the trick Yeah I tried to avoid just using simple replace methods for those possible sql injections
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I am trying to get the placeholder attribute value and do a fadeIn with the label which has the placeholder value as a for value, but it's not working HTML: lt;html gt; lt;body gt;
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Learn how to create a placeholder for a 'select' box using HTML and JavaScript on Stack Overflow
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How to change the placeholder text in the 'Additional Comments (Customer Visible)' field? I have tried updating the out-of-the-box script for 'Modify Comments Label,' but it updates both the label and
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How to use placeholders in like clause in SQL query? Asked 3 years, 8 months ago Modified 3 years, 6 months ago Viewed 1k times
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